显示来自 SQL 的详细信息时出错 - error in displaying detail from SQL

error in displaying detail from SQL

本文关键字：详细信息出错 SQL 显示 | 更新日期: 2023-09-27

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a3629462/public_html/delete/single ad display.php on line 29

这是错误，这是部分

得到错误

// get value of id that sent from address bar 
$id=$_GET['id'];
$sql="SELECT * FROM $tbl_name WHERE id='$id'";
$result=mysql_query($sql);
$rows=mysql_fetch_array($result);
?>

确切的误差线为

$rows=mysql_fetch_array($result);

完整代码

<?php
    $host = ""; // Host name 
    $username = ""; // Mysql username 
    $password = ""; // Mysql password 
    $db_name = "a3629462_sell"; // Database name 
    $tbl_name = "forum_question"; // Table name 
// Connect to server and select databse.
    mysql_connect("$host", "$username", "$password") or die("cannot connect");
    mysql_select_db("$db_name") or die("cannot select DB");
// get value of id that sent from address bar 
    $id = $_GET['id'];
    $sql = "SELECT * FROM $tbl_name WHERE id='$id'";
    $result = mysql_query($sql);
    $rows = mysql_fetch_array($result);
    ?>

添加了显示错误的完整代码

删除：

$sql = "SELECT * FROM $tbl_name WHERE id='$id'";
    $result = mysql_query($sql);
    $rows = mysql_fetch_array($result);

试试这个

$query = mysql_query("SELECT * FROM `".$tbl_name."` WHERE id = ' ". $id ." '") or die(mysql_error());
  $rows = mysql_fetch_array($query);

我建议您尝试以下方法：

$sql="SELECT * FROM `" . $tbl_name . "` WHERE id='" . $id . "'";

如果仍然不起作用，请尝试

echo $sql;

并直接在数据库中运行它，以查看它是否成功，或者给出更详细的错误。