Question

$sql = "SELECT * FROM product_suplier LEFT JOIN furnizor ON furnizor.id_furnizor=product_suplier.id_furnizor WHERE product_suplier.id_product=$_GET[id])";
        $result = $conn->query($sql);
        while($row_furnizor = $result->fetch_assoc()) {
        echo $row_furnizor['name'].' - adresa: '.$row_furnizor['adresa']; }

PHPMyadmin结果是：

2 ......... | ... 2 .................. | ... 1 ......... .......... | .... 1 ............... | ...凯文。| ...阿姆斯特丹
3 ......... | ... 2 .................. | ... 2 ............ ...... | ...... 2 ............... | ......关心...... | ......伦敦

错误：致命错误：在非对象上调用成员函数fetch_assoc（）

Answer 1

主要问题可能是缺少GET变量名称周围的引号 - 尽管没有逻辑测试来决定是否处理记录集。您可能还发现在连接表时更容易分配表别名，因为它使sql更容易阅读IMO。

<?php

    /* only proceed if there is an `id` ~ the $_GET variable required quotes */

    $id=isset( $_GET['id'] ) && !empty( $_GET['id'] ) ? strip_tags( trim( $_GET['id'] ) ): false;

    if( $id ){

        /* prepare the sql ( not prepared statement )*/
        $sql = "select * from 
            `product_suplier` p
            left outer join `furnizor` f on f.`id_furnizor`=p.`id_furnizor`
            where p.`id_product`='{$id}'";
        /* Assign a result to the query */
        $result = $conn->query( $sql );

        /* logical test on the query result - only proceed if it succeeded */
        if( $result ){

            /* Personal preference: using object notation rather than array */
            while( $rs = $result->fetch_object() ) {
                echo $rs->name.' - adresa: '.$rs->adresa; 
            }
        }
    }
?>

Answer 2

将此行更新为：

$sql = "SELECT * FROM product_suplier LEFT JOIN furnizor ON
furnizor.id_furnizor=product_suplier.id_furnizor WHERE 
product_suplier.id_product=".mysqli_escape_string($_GET['id']);

以无效方式调用$_GET数组值，同时删除)，当您从用户获取数据时应使用mysql_escape_string函数以防止SQL注入。

MySQL左连接 - PHP调用非对象

2 个答案: