php - 将数组传递给sql查询不起作用

Question

这个问题在这里已经有了答案:





mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource

(31 个回答)


8年前关闭。

<?php
require_once("dbdata.php");
if(mysql_connect($server_name,$Db_user,$Db_pass))
 {
if(mysql_select_db($Database_name))
   {
     $display_scores = "SELECT DISTINCT User_id, Quiz_id, Parent_Category,Category_Name, Score FROM custom_question_details where User_id = 3";
     $results = mysql_query($display_scores);
     $scores = array();
     $s_p = array();
     while($rows = mysql_fetch_assoc($results))
     {
        $scores[] = $rows; 
        $s_p [] = $rows['Parent_Category'];
     }
     $imp = implode(',',$s_p);
     $child_sql = "SELECT DISTINCT Category_Name FROM custom_question_details WHERE Parent_Category IN (".$imp.")";
     $child_result = mysql_query($child_sql);
     $child_array = array();
     while($fetch_child_results = mysql_fetch_assoc($child_result))
     {
         $child_array[] = $fetch_child_results;
     }
     print_r($child_array);
   }
 }
?>

我收到此错误

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in line ...

通过 $imp至 $child_sql询问。但是，如果我静态给出一些字符串，例如 'Sample'，它在 db 中而不是 $imp变量 i 获得结果。传递有什么问题 $imp传递查询的变量。

Answer 1

您必须在 IN 中引用值条款

尝试这个

$s_p [] = "'".$rows['Parent_Category']."'";

代替

$s_p [] = $rows['Parent_Category'];

现在使用它

$imp = implode(',',$s_p);
     $child_sql = "SELECT DISTINCT Category_Name FROM custom_question_details WHERE Parent_Category IN (".$imp.")";