在PHP中，当订购多个项目时，如何将这些项目的数量减少1？

这是我使用的代码；问题是只有1项 数量可以一次递减；不能输入多个项目数量 减少

 <?php


    $dbhost = "localhost"; 
    $dbuser = "root";
    $dbpass = "";
    $dbdb = "yumyum";


    $connect = mysql_connect($dbhost, $dbuser, $dbpass) or die("connection error");


    mysql_select_db($dbdb) or die("database selection error");


    $id = $_POST['id'];


    $query1=mysql_query("SELECT Quantity,id FROM `yumyum`.`food` where  `food`.`id` LIKE $id");
    $rows = array();

    while($r = mysql_fetch_assoc($query1)) {

        $output = $r['Quantity'];


            echo $output;
            $query2=mysql_query("UPDATE food SET Quantity = Quantity - 1 where  `food`.`id` LIKE ".$r["id"]);
    }

    ?>

你能试试这个吗

$query2 = mysql_query("UPDATE food SET Quantity = Quantity - 1 where  `food`.`id` = '".$r["id"]."' ");

或

---

因为它在$id=$\u POST['id']之后一次发送一个id

您只需执行以下操作：

mysql_query("UPDATE food SET Quantity = Quantity - 1 where  `food`.`id` = $id");

---

并删除它下面的所有其他行。

在记录查询后检查记录的编号。例如，对于您的代码

  <?php
    $dbhost = "localhost"; 
    $dbuser = "root";
    $dbpass = "";
    $dbdb = "yumyum";


    $connect = mysql_connect($dbhost, $dbuser, $dbpass) or die("connection error");

    mysql_select_db($dbdb) or die("database selection error");

    $id = $_POST['id'];

    $query1=mysql_query("SELECT Quantity,id FROM `yumyum`.`food` where  `food`.`id` LIKE $id");
    echo mysql_num_rows($query1); // this will show you count of records by above select query make sure you have more than one item count to update multiple record. 
    $rows = array();

    while($r = mysql_fetch_assoc($query1)) {

        $output = $r['Quantity'];
        echo $output;
        $query2=mysql_query("UPDATE food SET Quantity = Quantity - 1 where  `food`.`id` LIKE ".$r["id"]);
}

?>

---

根据你的数据库结构尝试使用通配符。

@Loz Cherone你能帮忙吗me@AndreiNicusan你能帮忙吗？你的语句的问题是IN子句的SELECT只能有1个字段。因此，将$subQuery select更改为select id FROM…@Qambar Raza没有说任何东西它不会更新多个项目的数量如果您的id是整数，我认为我们应该使用它作为子查询select id FROM yumyum.food`where food.id=$id`我如何相应地更改此部分？而$r=mysql\u fetch\u assoc$query1{$output=$r['Quantity']；//将返回的数据编码为JSON格式echo$output；$query2=mysql\u queryUPDATE food SET Quantity=Quantity-1 where food.id LIKE.$r[id]@QambarRaza有任何mor建议吗？您在输出中得到了多少？在打印mysql_num_行时，如果您可以粘贴它们，则很容易解决问题。注意：第1行的C:\wamp\www\yumyum\reserve.php中未定义的索引：id警告：mysql_num_行希望参数1是资源，第15行的C:\wamp\www\yumyum\reserve.php中给出的空值我已更新了我的co有一个拼写错误，请现在尝试…并分享结果

  <?php
    $dbhost = "localhost"; 
    $dbuser = "root";
    $dbpass = "";
    $dbdb = "yumyum";


    $connect = mysql_connect($dbhost, $dbuser, $dbpass) or die("connection error");

    mysql_select_db($dbdb) or die("database selection error");

    $id = $_POST['id'];

    $query1=mysql_query("SELECT Quantity,id FROM `yumyum`.`food` where  `food`.`id` LIKE $id");
    echo mysql_num_rows($query1); // this will show you count of records by above select query make sure you have more than one item count to update multiple record. 
    $rows = array();

    while($r = mysql_fetch_assoc($query1)) {

        $output = $r['Quantity'];
        echo $output;
        $query2=mysql_query("UPDATE food SET Quantity = Quantity - 1 where  `food`.`id` LIKE ".$r["id"]);
}

?>

    $query1 = mysql_query("SELECT Quantity,id FROM `yumyum`.`food` where  `food`.`id` LIKE '%".$id."%'");