blocks|key|1104974|text|以防你不得不在没有库帮助的情况下做到这一点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1104975|("00000000"+%2B+"Apple").substring("Apple".length())|code-block|syntax|javascript|1104976|(只要字符串长度不超过8个字符，就可以工作。)|1104977|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

In case you have to do it without the help of a library:

<pre><code>("00000000" + "Apple").substring("Apple".length())
</code></pre>

(Works, as long as your String isn't longer than 8 chars.)

blocks|key|1104839|text|public+class+LeadingZerosExample+{
++++public+static+void+main(String[]+args)+{
+++++++int+number+=+1500;

+++++++//+String+format+below+will+add+leading+zeros+(the+%250+syntax)+
+++++++//+to+the+number+above.+
+++++++//+The+length+of+the+formatted+string+will+be+7+characters.

+++++++String+formatted+=+String.format("%2507d",+number);

+++++++System.out.println("Number+with+leading+zeros:+"+%2B+formatted);
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1104840|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public class LeadingZerosExample {
 public static void main(String[] args) {
 int number = 1500;

 // String format below will add leading zeros (the %0 syntax) 
 // to the number above. 
 // The length of the formatted string will be 7 characters.

 String formatted = String.format("%07d", number);

 System.out.println("Number with leading zeros: " + formatted);
 }
}
</code></pre>

blocks|key|889887|text|+StringUtils.leftPad(yourString,+8,+'0');|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|889888|这是来自commons-lang的。See+javadoc|unstyled|offset|length|889889|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/lang/|1|https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#leftPad-java.lang.String-int-char-^0|0|4|C|0|I|B|1|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@]|9|@$G|U|H|V|1|W]|$G|X|H|Y|1|Z]]|A|$]]|$1|I|3|-4|5|F|7|10|8|@]|9|@]|A|$]]]|J|$K|$5|L|M|N|A|$O|P]]|Q|$5|L|M|N|A|$O|R]]]]

<pre><code> StringUtils.leftPad(yourString, 8, '0');
</code></pre>

This is from <a href="http://commons.apache.org/lang/" rel="noreferrer">commons-lang</a>. <a href="https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#leftPad-java.lang.String-int-char-" rel="noreferrer">See javadoc</a>

blocks|key|890219|text|我相信这就是他真正想要的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|890220|String.format("%250"%2B+(8+-+"Apple".length()+)%2B"d%25s",0+,"Apple");+|code-block|syntax|javascript|890221|输出：|890222|000Apple|890223|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

This is what he was really asking for I believe:

<pre><code>String.format("%0"+ (8 - "Apple".length() )+"d%s",0 ,"Apple"); 
</code></pre>

output:

<pre><code>000Apple
</code></pre>

blocks|key|1105065|text|您可以使用另一个答案中使用的String.format方法来生成包含0的字符串。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1105066|String.format("%250"%2Blength%2B"d",0)|code-block|syntax|javascript|1105067|这可以通过动态调整格式字符串中前导0的数量来解决您的问题：|1105068|public+String+leadingZeros(String+s,+int+length)+{
+++++if+(s.length()+>=+length)+return+s;
+++++else+return+String.format("%250"+%2B+(length-s.length())+%2B+"d%25s",+0,+s);
}|1105069|它仍然是一个混乱的解决方案，但它的优点是您可以使用整数参数指定结果字符串的总长度。|1105070|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You can use the String.format method as used in another answer to generate a string of 0's,

<pre><code>String.format("%0"+length+"d",0)
</code></pre>

This can be applied to your problem by dynamically adjusting the number of leading 0's in a format string:

<pre><code>public String leadingZeros(String s, int length) {
 if (s.length() &gt;= length) return s;
 else return String.format("%0" + (length-s.length()) + "d%s", 0, s);
}
</code></pre>

It's still a messy solution, but has the advantage that you can specify the total length of the resulting string using an integer argument.

blocks|key|1105219|text|使用Guava的Strings实用程序类：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1105220|Strings.padStart("Apple",+8,+'0');|code-block|syntax|javascript|1105221|entityMap|0|LINK|mutability|MUTABLE|url|http://google.github.io/guava/releases/19.0/api/docs/com/google/common/base/Strings.html^0|8|7|8|7|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Using Guava's <a href="http://google.github.io/guava/releases/19.0/api/docs/com/google/common/base/Strings.html" rel="noreferrer"><code>Strings</code></a> utility class:

<pre><code>Strings.padStart("Apple", 8, '0');
</code></pre>

blocks|key|890275|text|您可以使用以下命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|890276|org.apache.commons.lang.StringUtils.leftPad("Apple",+8,+"0")|code-block|syntax|javascript|890277|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can use this:

<pre><code>org.apache.commons.lang.StringUtils.leftPad("Apple", 8, "0")
</code></pre>

blocks|key|1105261|text|我遇到过类似的情况，我使用了这个；它非常简洁，您不必处理长度或其他库。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1105262|String+str+=+String.format("%258s","Apple");
str+=+str.replace('+','0');|code-block|syntax|javascript|1105263|简单整洁。字符串格式返回"+++Apple"，因此在将空格替换为零之后，它会给出所需的结果。|offset|length|style|CODE|1105264|entityMap^0|0|0|C|A|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I've been in a similar situation and I used this; It is quite concise and you don't have to deal with length or another library.

<pre><code>String str = String.format("%8s","Apple");
str = str.replace(' ','0');
</code></pre>

Simple and neat. String format returns <code>" Apple"</code> so after replacing space with zeros, it gives the desired result.

blocks|key|890092|text|String+input+=+"Apple";
StringBuffer+buf+=+new+StringBuffer(input);

while+(buf.length()+<+8)+{
++buf.insert(0,+'0');
}

String+output+=+buf.toString();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|890093|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>String input = "Apple";
StringBuffer buf = new StringBuffer(input);

while (buf.length() &lt; 8) {
 buf.insert(0, '0');
}

String output = buf.toString();
</code></pre>

blocks|key|1104896|text|使用Apache+Commons+StringUtils.leftPad+(或者查看代码来创建自己的函数)。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1104897|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/lang/api-2.5/org/apache/commons/lang/StringUtils.html#leftPad%2528java.lang.String,%2520int,%2520char%2529^0|2|Y|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Use <a href="http://commons.apache.org/lang/api-2.5/org/apache/commons/lang/StringUtils.html#leftPad%28java.lang.String,%20int,%20char%29" rel="noreferrer">Apache Commons StringUtils.leftPad</a> (or look at the code to make your own function).

blocks|key|1105556|text|您可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1105557|String.format("%2508d",+"Apple");|code-block|syntax|javascript|1105558|这似乎是最简单的方法，并且不需要任何外部库。|1105559|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You can use:
<pre><code>String.format(&quot;%08d&quot;, &quot;Apple&quot;);
</code></pre>
It seems to be the simplest method and there is no need of any external library.

blocks|key|890456|text|在Java中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|890457|String+zeroes="00000000";
String+apple="apple";

String+result=zeroes.substring(apple.length(),zeroes.length())%2Bapple;|code-block|syntax|javascript|890458|在Scala中：|890459|"Apple".foldLeft("00000000"){(ac,e)=>ac.tail%2Be}|890460|您还可以在Java+8中探索一种使用streams和reduce来完成此任务的方法(类似于我使用Scala完成此任务的方法)。它与所有其他解决方案都有点不同，我特别喜欢它。|890461|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

In Java:

<pre><code>String zeroes="00000000";
String apple="apple";

String result=zeroes.substring(apple.length(),zeroes.length())+apple;
</code></pre>

In Scala:

<pre><code>"Apple".foldLeft("00000000"){(ac,e)=&gt;ac.tail+e}
</code></pre>

You can also explore a way in Java 8 to do it using streams and reduce (similar to the way I did it with Scala). It's a bit different to all the other solutions and I particularly like it a lot.

blocks|key|890021|text|public+class+PaddingLeft+{
++++public+static+void+main(String[]+args)+{
++++++++String+input+=+"Apple";
++++++++String+result+=+"00000000"+%2B+input;
++++++++int+length+=+result.length();
++++++++result+=+result.substring(length+-+8,+length);
++++++++System.out.println(result);
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|890022|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public class PaddingLeft {
 public static void main(String[] args) {
 String input = "Apple";
 String result = "00000000" + input;
 int length = result.length();
 result = result.substring(length - 8, length);
 System.out.println(result);
 }
}
</code></pre>

blocks|key|1105022|text|你可能得照顾好埃奇卡斯。这是一个泛型方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1105023|public+class+Test+{
++++public+static+void+main(String[]+args){
++++++++System.out.println(padCharacter("0",8,"hello"));
++++}
++++public+static+String+padCharacter(String+c,+int+num,+String+str){
++++++++for(int+i=0;i<=num-str.length()%2B1;i%2B%2B){str+=+c%2Bstr;}
++++++++return+str;
++++}
}|code-block|syntax|javascript|1105024|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You may have to take care of edgecase. This is a generic method.

<pre><code>public class Test {
 public static void main(String[] args){
 System.out.println(padCharacter("0",8,"hello"));
 }
 public static String padCharacter(String c, int num, String str){
 for(int i=0;i&lt;=num-str.length()+1;i++){str = c+str;}
 return str;
 }
}
</code></pre>

blocks|key|890098|text|public+static+void+main(String[]+args)
{
++++String+stringForTest+=+"Apple";
++++int+requiredLengthAfterPadding+=+8;
++++int+inputStringLengh+=+stringForTest.length();
++++int+diff+=+requiredLengthAfterPadding+-+inputStringLengh;
++++if+(inputStringLengh+<+requiredLengthAfterPadding)
++++{
++++++++stringForTest+=+new+String(new+char[diff]).replace("\0",+"0")%2B+stringForTest;
++++}
++++System.out.println(stringForTest);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|890099|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public static void main(String[] args)
{
 String stringForTest = "Apple";
 int requiredLengthAfterPadding = 8;
 int inputStringLengh = stringForTest.length();
 int diff = requiredLengthAfterPadding - inputStringLengh;
 if (inputStringLengh &lt; requiredLengthAfterPadding)
 {
 stringForTest = new String(new char[diff]).replace("\0", "0")+ stringForTest;
 }
 System.out.println(stringForTest);
}
</code></pre>

blocks|key|1105332|text|public+static+String+lpad(String+str,+int+requiredLength,+char+padChar)+{
++++if+(str.length()+>+requiredLength)+{
++++++++return+str;
++++}+else+{
++++++++return+new+String(new+char[requiredLength+-+str.length()]).replace('\0',+padChar)+%2B+str;
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1105333|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public static String lpad(String str, int requiredLength, char padChar) {
 if (str.length() &gt; requiredLength) {
 return str;
 } else {
 return new String(new char[requiredLength - str.length()]).replace('\0', padChar) + str;
 }
}
</code></pre>

blocks|key|1105436|text|是否有人尝试过这个纯Java解决方案(没有SpringUtils)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1105437|//decimal+to+hex+string+1=>+01,+10=>0A,..
String.format("%251$2s",+Integer.toString(1,16)+).replace("+","0");
//reply+to+original+question,+string+with+leading+zeros.+
//first+generates+a+10+char+long+string+with+leading+spaces,+and+then+spaces+are
//replaced+by+a+zero+string.+
String.format("%251$10s",+"mystring"+).replace("+","0");|code-block|syntax|javascript|1105438|不幸的是，只有当字符串中没有空格时，此解决方案才有效。|1105439|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Did anyone tried this pure Java solution (without SpringUtils): 

<pre><code>//decimal to hex string 1=&gt; 01, 10=&gt;0A,..
String.format("%1$2s", Integer.toString(1,16) ).replace(" ","0");
//reply to original question, string with leading zeros. 
//first generates a 10 char long string with leading spaces, and then spaces are
//replaced by a zero string. 
String.format("%1$10s", "mystring" ).replace(" ","0");
</code></pre>

Unfortunately this solution works only if you do not have blank spaces in a string.

blocks|key|890590|text|使用String::repeat方法的解决方案(Java+11)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|890591|String+str+=+"Apple";
String+formatted+=+"0".repeat(8+-+str.length())+%2B+str;|code-block|syntax|javascript|890592|如果需要，将8更改为另一个数字或将其参数化|890593|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Solution with method String::repeat (Java 11)
<pre><code>String str = &quot;Apple&quot;;
String formatted = &quot;0&quot;.repeat(8 - str.length()) + str;
</code></pre>
If needed change 8 to another number or parameterize it

blocks|key|890741|text|我喜欢Pad+a+String+with+Zeros的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|890742|String.format("%251$"+%2B+length+%2B+"s",+inputString).replace('+',+'0');|code-block|syntax|javascript|890743|长度=+"8“，inputString+=”苹果“|890744|entityMap|0|LINK|mutability|MUTABLE|url|https://www.baeldung.com/java-pad-string^0|3|N|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

I like the solution from <a href="https://www.baeldung.com/java-pad-string" rel="nofollow noreferrer">Pad a String with Zeros</a>
<pre><code>String.format(&quot;%1$&quot; + length + &quot;s&quot;, inputString).replace(' ', '0');
</code></pre>
with length = &quot;8&quot; and inputString = &quot;Apple&quot;

blocks|key|1105109|text|这是快速的&适用于任何长度。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1105110|public+static+String+prefixZeros(String+value,+int+len)+{
++++char[]+t+=+new+char[len];
++++int+l+=+value.length();
++++int+k+=+len-l;
++++for(int+i=0;i<k;i%2B%2B)+{+t[i]='0';+}
++++value.getChars(0,+l,+t,+k);
++++return+new+String(t);
}|code-block|syntax|javascript|1105111|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is fast &amp; works for whatever length.

<pre><code>public static String prefixZeros(String value, int len) {
 char[] t = new char[len];
 int l = value.length();
 int k = len-l;
 for(int i=0;i&lt;k;i++) { t[i]='0'; }
 value.getChars(0, l, t, k);
 return new String(t);
}
</code></pre>

blocks|key|890264|text|当大多数in字符串都有8个字符时，可以比Chris+Lercher+answer更快|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|890265|int+length+=+in.length();
return+length+==+8+?+in+:+("00000000"+%2B+in).substring(length);|code-block|syntax|javascript|890266|在我的情况下，在我的机器上快了1/8。|890267|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/4051991/4891683^0|K|K|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

Can be faster then <a href="https://stackoverflow.com/a/4051991/4891683">Chris Lercher answer</a> when most of in String have exacly 8 char

<pre><code>int length = in.length();
return length == 8 ? in : ("00000000" + in).substring(length);
</code></pre>

in my case on my machine 1/8 faster.

blocks|key|890395|text|下面是我用来预先填充字符串的简单的API-less+"readable+script“版本。(简单、可读和可调)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|890396|while(str.length()+<+desired_length)
++str+=+'0'%2Bstr;|code-block|syntax|javascript|890397|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is the simple API-less "readable script" version I use for pre-padding a string. (Simple, Readable, and Adjustable).

<pre><code>while(str.length() &lt; desired_length)
 str = '0'+str;
</code></pre>

blocks|key|890003|text|它不是很漂亮，但它是有效的。如果您有apache+commons访问权限，我建议您使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|890004|if+(val.length()+<+8)+{
++for+(int+i+=+0;+i+<+val+-+8;+i%2B%2B)+{
++++val+=+"0"+%2B+val;
++}
}|code-block|syntax|javascript|890005|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

It isn't pretty, but it works. If you have access apache commons i would suggest that use that

<pre><code>if (val.length() &lt; 8) {
 for (int i = 0; i &lt; val - 8; i++) {
 val = "0" + val;
 }
}
</code></pre>

Here is the String, for example:

<pre><code>"Apple"
</code></pre>

and I would like to add zero to fill in 8 chars:

<pre><code>"000Apple"
</code></pre>

How can I do so?

How to format a Java string with leading zero?

以下是字符串，例如："Apple"我想添加0来填充8个字符："000Apple"我如何做到这一点呢？

问如何格式化带有前导零的Java字符串？
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问如何格式化带有前导零的Java字符串？
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