blocks|key|3162158|text|$var+=+"partial%25";
$stmt+=+$dbh->prepare("select+wrd+from+tablename+WHERE+wrd+LIKE+:partial");
$stmt->bindParam(":partial",+$var);
$stmt->execute();+//+or+$stmt->execute(array(':partial'+=>+$var));+without+
++++++++++++++++++//+first+calling+bindParam()
$rs+=+$stmt->fetchAll();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3162159|使用问号参数：|unstyled|3162160|$stmt+=+$dbh->prepare('select+wrd+from+tablename+WHERE+wrd+LIKE+?');
$stmt->execute(array('partial%25'));
$rs+=+$stmt->fetchAll();|3162161|http://www.php.net/manual/en/pdo.prepare.php|offset|length|3162162|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|0|18|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|V|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|W|8|@]|9|@$K|X|L|Y|1|Z]]|A|$]]|$1|M|3|-4|5|F|7|10|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|J]]]]

<pre><code>$var = "partial%";
$stmt = $dbh-&gt;prepare("select wrd from tablename WHERE wrd LIKE :partial");
$stmt-&gt;bindParam(":partial", $var);
$stmt-&gt;execute(); // or $stmt-&gt;execute(array(':partial' =&gt; $var)); without 
 // first calling bindParam()
$rs = $stmt-&gt;fetchAll();
</code></pre>

Using question mark parameters:

<pre><code>$stmt = $dbh-&gt;prepare('select wrd from tablename WHERE wrd LIKE ?');
$stmt-&gt;execute(array('partial%'));
$rs = $stmt-&gt;fetchAll();
</code></pre>

<a href="http://www.php.net/manual/en/pdo.prepare.php" rel="noreferrer">http://www.php.net/manual/en/pdo.prepare.php</a>

blocks|key|3162180|text|你也可以说：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3162181|SELECT+wrd+FROM+tablename+WHERE+wrd+LIKE+CONCAT(:partial,+'%25')|code-block|syntax|javascript|3162182|在MySQL末端连接字符串，在这种情况下没有任何特殊的原因。|3162183|如果要查找的部分wrd本身可以包含百分号或下划线字符(因为它们对LIKE操作符有特殊意义)或反斜杠(根据ANSI+SQL标准，MySQL在LIKE操作符 - 中错误地使用了它作为另一层转义层)，事情就会变得更复杂。|offset|length|style|CODE|3162184|希望这不会影响你，但如果你真的需要正确的情况，这里有一个混乱的解决方案：|3162185|$stmt=+$db->prepare("SELECT+wrd+FROM+tablename+WHERE+wrd+LIKE+:term+ESCAPE+'%2B'");
$escaped=+str_replace(array('%2B',+'%25',+'_'),+array('%2B%2B',+'%2B%25',+'%2B_'),+$var);
$stmt->bindParam(':term',+$escaped);|3162186|entityMap^0|0|0|0|8|3|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@$K|Y|L|Z|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

You could also say:

<pre><code>SELECT wrd FROM tablename WHERE wrd LIKE CONCAT(:partial, '%')
</code></pre>

to do the string joining at the MySQL end, not that there's any particular reason to in this case.

Things get a bit more tricky if the partial <code>wrd</code> you are looking for can itself contain a percent or underscore character (since those have special meaning for the LIKE operator) or a backslash (which MySQL uses as another layer of escaping in the LIKE operator — incorrectly, according to the ANSI SQL standard).

Hopefully that doesn't affect you, but if you do need to get that case right, here's the messy solution:

<pre><code>$stmt= $db-&gt;prepare("SELECT wrd FROM tablename WHERE wrd LIKE :term ESCAPE '+'");
$escaped= str_replace(array('+', '%', '_'), array('++', '+%', '+_'), $var);
$stmt-&gt;bindParam(':term', $escaped);
</code></pre>

blocks|key|1753999|text|谁写了答案(可能是karim79)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1754000|1754001|+$var+="partial%25“$stmt+=$dbh->prepare("select+wrd+from+tablename+WHERE+wrd+LIKE+:partial")+$stmt->bindParam(":partial"，$var)+$stmt->execute()；//或$stmt->execute(array(‘：partial’=>$var))；不//首先调用bindParam()+$rs+=$stmt->fetchAll()；|blockquote|1754002|1754003|使用问号参数：|1754004|$stmt+=$dbh->prepare('select+wrd+from+tablename+WHERE+wrd+LIKE+?')；$stmt->execute(array('partial%25'))；$rs+=$stmt->fetchAll()；|1754005|1754006|非常感谢他。我正在搜索代码，看到了很多示例，但我无法解决我的问题。这次我成功地做到了这一点。我使用了代码的“使用问号参数：”部分。|1754007|对于其他人的帮助，如果您想从变量中检索值，可以将代码更改为|1754008|$stmt->execute(array($variable.'%25'));|code-block|syntax|javascript|1754009|而不是|1754010|$stmt->execute(array('partial%25'));|1754011|因为“partial”这个词是在答案中指定的，不能更改。非常感谢。|1754012|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|13|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|14|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|16|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|17|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|18|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|19|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1A|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|1B|8|@]|9|@]|A|$S|T]]|$1|U|3|V|5|6|7|1C|8|@]|9|@]|A|$]]|$1|W|3|X|5|R|7|1D|8|@]|9|@]|A|$S|T]]|$1|Y|3|Z|5|6|7|1E|8|@]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]]|11|$]]

Who has written the answare (may be karim79):

<blockquote>
<pre><code>$var ="partial%"
$stmt =$dbh-&gt;prepare("select wrd from tablename WHERE wrd LIKE :partial")
$stmt-&gt;bindParam(":partial",$var)
$stmt-&gt;execute(); //or$stmt-&gt;execute(array(':partial'=&gt;$var)); without 
 // first calling bindParam()
$rs =$stmt-&gt;fetchAll();
</code></pre>
 
 Using question mark parameters:

<pre><code>$stmt =$dbh-&gt;prepare('select wrd from tablename WHERE wrd LIKE ?');
$stmt-&gt;execute(array('partial%'));
$rs =$stmt-&gt;fetchAll();
</code></pre>
</blockquote>

Many thanks to him. I was searching for the code &amp; saw many examples, but i couldn't resolve my issue. This time I have succeed to do it. I used the 'Using question mark parameters:' section of the code. 

For others help, if you want to retrieve the value from a variable you may change the code to 

<pre><code>$stmt-&gt;execute(array($variable.'%'));
</code></pre>

instead of

<pre><code>$stmt-&gt;execute(array('partial%'));
</code></pre>

Because the word 'partial' is specified in the answer and can't be changed. Thanks a lot.

blocks|key|396538|text|这就是你应该怎么做|type|unstyled|depth|inlineStyleRanges|entityRanges|data|396539|bindValue(':partial',+'%25'+.+$_GET['partial']+.+'%25');|code-block|syntax|javascript|396540|谢谢,|396541|查询|396542|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

This is how you should do it

<pre><code>bindValue(':partial', '%' . $_GET['partial'] . '%');
</code></pre>

Thanks, 

Qwerty

blocks|key|396250|text|下面的代码只显示了数据库中的第一个关键字！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|396251|"SELECT+wrd+FROM+tablename+WHERE+wrd+LIKE+CONCAT(:partial,+'%25')"|code-block|syntax|javascript|396252|如果要从数据库中搜索所有关键字，请尝试此方法|396253|"SELECT+wrd+FROM+tablename+WHERE+wrd+LIKE+:partial";
$stmt->execute(array(':partial'=>'%25'.$YourVarHere.'%25'));|396254|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|$]]

The below code it shows only the first keywords in the database! 

<pre><code>"SELECT wrd FROM tablename WHERE wrd LIKE CONCAT(:partial, '%')"
</code></pre>

Try this one if you want to search all the keywords from the database

<pre><code>"SELECT wrd FROM tablename WHERE wrd LIKE :partial";
$stmt-&gt;execute(array(':partial'=&gt;'%'.$YourVarHere.'%'));
</code></pre>

blocks|key|396582|text|您可以在准备好的语句之前使用addcslashes。我在mysql上进行了测试。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|396583|$value+=+addcslashes($value,+'%25');
$stmt+=+$db->prepare('select+*+from+products+where+description+like+?');
$stmt->execute(["$value%25"]);|code-block|syntax|javascript|396584|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 addcslashes before prepared statement. I tested on mysql.

<pre><code>$value = addcslashes($value, '%');
$stmt = $db-&gt;prepare('select * from products where description like ?');
$stmt-&gt;execute(["$value%"]);
</code></pre>

blocks|key|3162446|text|我认为公认的答案(+@bobince)可以简化一点。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3162447|您可以将它简化为这样的形式，以处理参数中的下划线、百分比等，但仍然使用部分%25匹配like查询：|3162448|$stmt+=+$dbh->prepare("select+wrd+from+tablename+WHERE+wrd+LIKE+:partial");
$stmt->execute([":partial"+=>+addcslashes($value,+'_%25')+.+"%25"]);
$rows+=+$stmt->fetchAll();|code-block|syntax|javascript|3162449|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I think the accepted answer (by @bobince) can be simplified a bit.

You can reduce it to something like this to handle underscore, percentage, etc in the param but still match the LIKE query with partial%:

<pre><code>$stmt = $dbh-&gt;prepare("select wrd from tablename WHERE wrd LIKE :partial");
$stmt-&gt;execute([":partial" =&gt; addcslashes($value, '_%') . "%"]);
$rows = $stmt-&gt;fetchAll();
</code></pre>

In this query 

<pre><code>select wrd from tablename WHERE wrd LIKE '$partial%'
</code></pre>

I'm trying to bind the variable <code>'$partial%'</code> with PDO. Not sure how this works with the <code>%</code> at the end. 

Would it be 

<pre><code>select wrd from tablename WHERE wrd LIKE ':partial%'
</code></pre>

where <code>:partial</code> is bound to <code>$partial="somet"</code> 

or would it be 

<pre><code>select wrd from tablename WHERE wrd LIKE ':partial'
</code></pre>

where <code>:partial</code> is bound to <code>$partial="somet%"</code>

or would it be something entirely different?

mysql PDO how to bind LIKE

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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在此查询中select wrd from tablename WHERE wrd LIKE '$partial%'我正在尝试将变量'$partial%'与PDO绑定。最后，我不确定这与%是如何工作的。会不会是select wrd from tablename WHERE wrd LIKE ':partial%':par...

问mysql PDO如何像这样绑定
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问mysql PDO如何像这样绑定EN