本文介绍了如何使用LIKE子句询问准备查询的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

请看下面的代码这与mysqli方法工作正常,但很容易sql注入所以我想使用准备但是喜欢我不能使用它



$ t = strtolower($ _ POST ['e']);

$ search_exploded = explode(,$ t);

$ construct ='';

foreach($ search_exploded as $ search_each){

$ construct。=AND title LIKE?;

}





$ query = $ conn-> prepare(SELECT * FROM vdo WHERE 1 $ construct);

$ query-> execute([%$ search_each%]);

$ found = $ query-> rowCount();

if($ found == 0){

echoNO Result Found;



}其他{

while($ row_id1 = $ query-> fetch(PDO :: FETCH_ASSOC)){

echo $ title = $ row_id1 ['title'];



}



我尝试了什么:



请看下面的代码这与mysqli方法工作正常但很容易sql注入所以我想使用准备但是为了LIKE我不能使用它
解决方案



please see the below code this works fine with mysqli method but prone to sql injection so i want to use prepare but for LIKE i cant use it

$t=strtolower($_POST['e']);
$search_exploded = explode ( " ", $t );
$construct = '';
foreach( $search_exploded as $search_each ) {
$construct .="AND title LIKE ? ";
}


$query=$conn->prepare( "SELECT * FROM vdo WHERE 1 $construct ");
$query->execute(["%$search_each%"]);
$found=$query->rowCount();
if($found == 0){
echo "NO Result Found";

}else{
while($row_id1=$query->fetch(PDO::FETCH_ASSOC)){
echo $title=$row_id1['title'];

}

What I have tried:

please see the below code this works fine with mysqli method but prone to sql injection so i want to use prepare but for LIKE i cant use it
解决方案




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09-06 03:23