警告:mysql_fetch_array()期望参数1是资源,

时间:2015-10-03 07:49:01

标签: php

**

这个问题已经有了答案:

mysql_fetch_array() expects parameter 1 to be resource (or mysqli_result), boolean given 31 answers

我正在尝试从我的数据库中写入数据,这个代码意味着要做,但我得到两个警告:mysql_fetch_array()期望参数1是资源,布尔给定'并且不知道为什么???? **

     <?php
    $connection = mysql_connect("localhost", "root", ""); 
// Establishing Connection with Server
    $db = mysql_select_db("burger machine", $connection);
 // Selecting Database

    //MySQL Query to read data

    $query = mysql_query("select * from add", $connection);



    while ($row = mysql_fetch_array($query)) {

    echo "<tr>";
    echo " <td> {$row['Addons_id']} </td>";
    echo " <td> <a href=\"Ao.php?update={$row['Addons_id']}\"> {$row['Addons_Name']} </a> </td> ";
    echo " <td> {$row['Addons_Price']} </td> ";
    echo " <td> {$row['Addons_Description']} </td>";
    echo "</tr>";

    }

1 个答案:

答案 0 :(得分:0)

使用mysqli因为mysql现已弃用,并且还容易进行sql注入。

使用mysqli_query($con , $query);

其中$query是用于获取数据的SQL查询,$con是您的连接资源变量。

旁注

mysql_query();中,只需将您的查询作为参数传递,而不是您的连接。

<?php
    $connection = mysqli_connect("localhost", "root", ""); 
// Establishing Connection with Server
    $db = mysqli_select_db( $connection , "burger machine" );
 // Selecting Database

    //MySQL Query to read data

    $query = mysqli_query( $connection , "select * from add" );



    while ($row = mysqli_fetch_array($query)) {

    echo "<tr>";
    echo " <td> {$row['Addons_id']} </td>";
    echo " <td> <a href=\"Ao.php?update={$row['Addons_id']}\"> {$row['Addons_Name']} </a> </td> ";
    echo " <td> {$row['Addons_Price']} </td> ";
    echo " <td> {$row['Addons_Description']} </td>";
    echo "</tr>";
    }