blocks|key|121966|text|LIKE+'%25searchq%25'"|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|121967|它会搜索像'searchq‘这样的字符串。如果您需要作为变量，请添加相应的美元符号|unstyled|121968|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>LIKE '%searchq%'"
</code></pre>

It searches for string like 'searchq' if you need to be the variable, add the respective dollar sign

blocks|key|124173|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|124174|<?php
mysql_connect("localhost","michael","xcA123sd")+or+die(mysql_error());
mysql_select_db("search_test")+or+die+("could+not+find+db");+
$output+='';
if+(isset($_get['search'])){
++++$searchq+=+$_get['search'];
}
$query+=+mysql_query("SELECT+*+FROM+members+WHERE+firstname+LIKE+$searchq"+)+or+die("could+not+search");
$count+=+mysql_num_rows($query);
if($count+==+0){
++++$output+=+'There+was+no+search+results+!';
}else{
++++while($row+=+mysql_fetch_array($query)){
++++++++$fname+=+$row['firstname'];
++++++++$output+.='<div>+'.$fname.'</div>';
++++}
}
?>

<!DOCTYPE+html+PUBLIC+"-//W3C//DTD+XHTML+1.0+Transitional//EN"+"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html+xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta+http-equiv="Content-Type"+content="text/html;+charset=utf-8"+/>
<title>search</title>
</head>
<body>
<form+action="index.php"+method="get">
<input+type="text"+name="search"+placeholder="search+for+members"/>+
<input+type="submit"+value=">>"/>
</form>
<?php+print("$output);?>
</body>
</html>|code-block|syntax|javascript|124175|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|$]]

Try this:

<pre><code>&lt;?php
mysql_connect("localhost","michael","xcA123sd") or die(mysql_error());
mysql_select_db("search_test") or die ("could not find db"); 
$output ='';
if (isset($_get['search'])){
 $searchq = $_get['search'];
}
$query = mysql_query("SELECT * FROM members WHERE firstname LIKE $searchq" ) or die("could not search");
$count = mysql_num_rows($query);
if($count == 0){
 $output = 'There was no search results !';
}else{
 while($row = mysql_fetch_array($query)){
 $fname = $row['firstname'];
 $output .='&lt;div&gt; '.$fname.'&lt;/div&gt;';
 }
}
?&gt;

&lt;!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"&gt;
&lt;html xmlns="http://www.w3.org/1999/xhtml"&gt;
&lt;head&gt;
&lt;meta http-equiv="Content-Type" content="text/html; charset=utf-8" /&gt;
&lt;title&gt;search&lt;/title&gt;
&lt;/head&gt;
&lt;body&gt;
&lt;form action="index.php" method="get"&gt;
&lt;input type="text" name="search" placeholder="search for members"/&gt; 
&lt;input type="submit" value="&gt;&gt;"/&gt;
&lt;/form&gt;
&lt;?php print("$output);?&gt;
&lt;/body&gt;
&lt;/html&gt;
</code></pre>

blocks|key|2827163|text|对于select语句，您需要：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2827164|$query+=+mysql_query("SELECT+*+FROM+members+WHERE+firstname+LIKE+'%25searchq%25'"+)+or+die("could+not+search");|code-block|syntax|javascript|2827165|它应该是：|2827166|$query+=+mysql_query("SELECT+*+FROM+members+WHERE+firstname+LIKE+'%25".$searchq."'%25"+)+or+die("could+not+search");|2827167|因为您搜索的是名为searchq的变量中的内容，而不是字符串searchq+:)|2827168|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|$]]

For your select statement, you have:

<pre><code>$query = mysql_query("SELECT * FROM members WHERE firstname LIKE '%searchq%'" ) or die("could not search");
</code></pre>

It should be:

<pre><code>$query = mysql_query("SELECT * FROM members WHERE firstname LIKE '%".$searchq."'%" ) or die("could not search");
</code></pre>

Because you're searching for the content inside the variable called searchq, not actually the string searchq :)

blocks|key|4939682|text|问题来自您代码的这一行|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4939683|$query+=+mysql_query("SELECT+*+FROM+members+WHERE+firstname+LIKE+'%25searchq%25'“)|4939684|变量searchq后面没有$|4939685|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

The problem come from this line of your code

$query = mysql_query("SELECT * FROM members WHERE firstname LIKE '%searchq%'" )

the variable searchq doesnot have $ at the back

blocks|key|122003|text|首先:你想|type|unstyled|depth|inlineStyleRanges|entityRanges|data|122004|$query+=+mysql_query("SELECT+*+FROM+members+WHERE+firstname+LIKE+'%25searchq%25'"+)+or+die("could+not+search");|code-block|syntax|javascript|122005|成为|122006|$query+=+mysql_query("SELECT+*+FROM+members+WHERE+firstname+LIKE+'%25$searchq%25'"+)+or+die("could+not+search");|122007|(注意附加的$)。|offset|length|style|CODE|122008|也就是说，您有一个很大的SQL注入问题:假设我“正常”运行了一次查询:这让我对列有了一个概念。现在我发布'+UNION+ALL+SELECT+correct_field_num+FROM+information_schema.TABLES+WHERE+NAME+LIKE+'%25作为我的搜索--这给了我你的表结构。使用posting+'+UNION+ALL+SELECT+correct_column_num+FROM+any_table_name+WHERE+'x'+LIKE+'%25，我可以读取任意表。|122009|请确保您使用了一种广为人知的技术来构造来自任何用户输入的安全查询。从不推荐使用的mysql_real_escape_string()到参数化查询都有不同程度的变化。|BOLD|122010|entityMap^0|0|0|0|0|6|1|0|1G|2E|4M|23|0|L|2|14|Q|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Z|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|11|8|@$M|12|N|13|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|14|8|@$M|15|N|16|O|P]|$M|17|N|18|O|P]]|9|@]|A|$]]|$1|S|3|T|5|6|7|19|8|@$M|1A|N|1B|O|U]|$M|1C|N|1D|O|P]]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|W|$]]

First of all: You want

<pre><code>$query = mysql_query("SELECT * FROM members WHERE firstname LIKE '%searchq%'" ) or die("could not search");
</code></pre>

to be 

<pre><code>$query = mysql_query("SELECT * FROM members WHERE firstname LIKE '%$searchq%'" ) or die("could not search");
</code></pre>

(mind the additional <code>$</code>). 

That said, you have a big SQL injection problem: Assume, I run the query "normally" once: This fives me an idea of the columns. Now I post <code>' UNION ALL SELECT correct_field_num FROM information_schema.TABLES WHERE NAME LIKE '%</code> as my search - this gives me your table structure. With posting <code>' UNION ALL SELECT correct_column_num FROM any_table_name WHERE 'x' LIKE '%</code> I can read an arbitrary table.

Make sure, you use one of the well-understood techniques to constuct a safe query from any user input. There is a spectrum from the deprecated <code>mysql_real_escape_string()</code> up to parameterized queries.

Hello I am having massive problems with this task for my assignment 

I have a database set up on xampp called search_test which has firstname and lastname as fields in it. I have set up a php form so when the user types in a name say Andre it returns all the andres in the database. There is a problem it keeps telling me there are no search results even though i know there is data in the database Here is the code supposed to be one php page called index.php 

<pre><code> &lt;?php
 mysql_connect("localhost","michael","xcA123sd") or die(mysql_error());
 mysql_select_db("search_test") or die ("could not find db"); 
 $output ='';
 if (isset ($_POST['search']));
 $searchq = $_POST['search'];
 $query = mysql_query("SELECT * FROM members WHERE firstname LIKE '%searchq%'" ) or die("could not search");
 $count = mysql_num_rows($query);
 if($count == 0){
 $output = 'There was no search results !';
 }else{
 while($row = mysql_fetch_array($query)){
 $fname = $row['firstname'];
 $output .='&lt;div&gt; '.$fname.'&lt;/div&gt;';
 }

 }
}


?&gt;

 &lt;!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"&gt;
&lt;html xmlns="http://www.w3.org/1999/xhtml"&gt;
&lt;head&gt;
&lt;meta http-equiv="Content-Type" content="text/html; charset=utf-8" /&gt;
&lt;title&gt;search&lt;/title&gt;
&lt;/head&gt;
&lt;body&gt;
&lt;form action="index.php" method="post"&gt;
&lt;input type="text" name="search" placeholder="search for members"/&gt; 
&lt;input type="submit" value="&gt;&gt;"/&gt;
&lt;/form&gt;
&lt;?php print("$output);?&gt;
&lt;/body
&lt;/html&gt; 
</code></pre>

for example i type andre in and i get the response 

There was no search results ! 

could someone please help

How to display MYSQL data into PHP search box

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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你好，我在这项任务中遇到了很大的问题我在xampp上建立了一个名为search_test的数据库，其中有名字和姓氏作为字段。我已经设置了一个php表单，这样当用户输入一个名字时，比如说Andre，它会返回数据库中的所有andres。有一个问题，它一直告诉我没有搜索结果，即使我知道数据库里有数据，这里的代码应该是一个叫做...

问如何在PHP搜索框中显示MYSQL数据
EN

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