Question

我刚刚编写了这段代码，它回应了它应该是什么，但是在echo语句之后它给了我错误 -

您的SQL语法有错误;检查手册对应于您的MySQL服务器版本，以便使用正确的语法靠近''第1行

为什么会这样？我该如何解决？

<?php
$myclasses = explode(',', $_SESSION['classlist']);  
$theirclasses = explode(',', $user_info['classlist']);
$common_classes = array_intersect($myclasses, $theirclasses);

if (count($common_classes) > 0) {     

    foreach ($common_classes as $class) {
        $classes = mysql_query("SELECT * FROM classes WHERE class_id = ".$class) or die(mysql_error());
        while($currentRow = mysql_fetch_array($classes)){
        echo $currentRow['class_name'];
        }
    }  
}
else {
}

?>

Answer 1

尝试使用quote包装查询：

$classes = mysql_query("SELECT * FROM classes WHERE class_id = '".$class."'") or die(mysql_error());

或使用PDO完全更改您的查询。因为，不推荐使用mysql_ *函数。

Answer 2

我将假设传递给查询时$ class变量出错。我在这种情况下通常做的是将SQL查询分配给字符串变量并转储它以立即测试整个查询。帮助我找出SQL语法错误或是否有任何不需要的字符。

<?php
$myclasses = explode(',', $_SESSION['classlist']);  
$theirclasses = explode(',', $user_info['classlist']);
$common_classes = array_intersect($myclasses, $theirclasses);

if (count($common_classes) > 0) {     

    foreach ($common_classes as $class) {
        $sql = "SELECT * FROM classes WHERE class_id = '{$class}'" ; // use {} inside double quotes
        var_dump($sql); // check out the what the query becomes 
        $classes = mysql_query($sql) or die(mysql_error());
        while($currentRow = mysql_fetch_array($classes)){
        echo $currentRow['class_name'];
        }
    }  
}
else {
}

Answer 3

如果$ class为空，则会收到该错误，该错误会使您的查询为SELECT * FROM classes WHERE class_id =无效。试试引用它。如果你引用它，至少你会得到SELECT * FROM classes WHERE class_id = ''

"SELECT * FROM classes WHERE class_id = '".$class."'";

PHP Works但是给出了SQL语法错误

3 个答案: