您的位置: 首页  >  技术问答  >  MySQL注意：未定义的索引：如何解决[重复]

MySQL注意：未定义的索引：如何解决[重复]

分类: 技术问答 • 2022-04-08 18:04:35
问题描述：

I am creating two level dropdown select box with PHP, MYSQL and JQuery. Everything seems fine but when my parent category doesn't have child or if I doesn't select child it is giving me error and I can't understand what this error means and how to resolve. I need you guyz help.

Error: Notice: Undefined index: f_child_cat in C:\xampp\htdocs\pathtofile\index.php on line 58 You selected 12 & You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

Here is my front end code

<?php if(isset($_POST['submit'])){
    $drop = $_POST['first_child_cat'];
    $f_child_cat = $_POST['f_child_cat'];
    echo "You selected ";
    echo $drop." & ".$f_child_cat;

    $query =    mysql_query("SELECT * FROM qa_categories WHERE parentid = $f_child_cat")
                or die(mysql_error());

                echo '<ul>';
                while($cats = mysql_fetch_array( $query ))
                    echo '<li>',$cats['title'],'</li>';

                echo '</ul>';
}
?>

</div>

此问题已经存在 这里有一个答案： p>

  • ”注意：未定义的变量“，”注意：未定义的索引“和”通知：未定义的偏移“使用PHP 28 answers span> li> ul> div>

    我正在使用PHP，MYSQL和JQuery创建两级下拉选择框。 一切似乎都很好，但是当我的父类别没有孩子或者我没有选择孩子时它会给我错误，我无法理解这个错误意味着什么以及如何解决。 我需要你的帮助。 p>

    错误： 注意：未定义的索引：第58行的C：\ xampp \ htdocs \ pathtofile \ index.php中的f_child_cat 您选择了12 ＆安培; 您的SQL语法有错误; 检查与MySQL服务器版本对应的手册，以便在第1行的''附近使用正确的语法 p> p>

    这是我的前端代码 p> \ n 

     ＆lt;？php if（isset（$ _ POST ['submit']））{
 $ drop = $ _POST ['first_child_cat']; 
 $ f_child_cat = $ _POST ['f_child_cat']  ; 
 echo“你选择”; 
 echo $ drop。“＆amp;”。$ f_child_cat; 
 
 $ query = mysql_query（“SELECT * FROM qa_categories WHERE parentid = $ f_child_cat”）
或者死（mysql_error）  （））; 
 
 echo'＆lt; ul＆gt;'; 
 while（$ cats = mysql_fetch_array（$ query））
 echo'＆lt; li＆gt;'，$ cats ['title']，'＆lt;  / li＆gt;'; 
 
 echo'＆lt; / ul＆gt;'; 
} 
？＆gt; 
  code>  pre> 
  div>

答

It means that $_POST['f_child_cat'] is not defined (doesn't exist). You should be using binded parameters to prevent against SQL injection and use mysqli_ or PDO functions.

Since this is part of your query, I would recommend that you wrap your query like so:

if( isset($_POST['f_child_cat']) && !empty($_POST['f_child_cat']) ) {
    ... call db
}
else {
    echo "You didn't select this"
}

答

$_POST['f_child_cat'] is not defined check if the value is present first before using it.

$f_child_cat = isset($_POST['f_child_cat'])?$_POST['f_child_cat']:false;

if ($f_child_cat){
    $f_child_cat = mysql_real_escape_string($f_child_cat); // always escape

    /// your query.
}

Note: mysql_* functions are depricated, use mysqli or pdo as an alternative.

相关推荐