根据条件更新表(循环时) [英] Update table based on condition (While Loop)

问题描述

所以我试图根据一个单参数来更新我的表:

So I am trying to update my table based on a singe parameter:

dateEntered字段必须为空白.

The dateEntered field must be blank.

我想随机选择50行，并将空白的ownerID字段更新为测试器"

And I want to randomly select 50 rows, and update the blank ownerID fields to "Tester"

这就是我所拥有的:

<?php
include("includes/constants.php");
include("includes/opendb.php");



$query = "SELECT * FROM contacts WHERE dateEntered='' ORDER BY RAND() LIMIT 50";
$result = mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_assoc($result)){
        $firstid = $row['id'];

        $query2 = mysql_query("UPDATE contacts 
                        SET ownerID = 'Tester' 
                   WHERE id = '$firstid'");

        $result2 = mysql_query($query2) or die(mysql_error());

        }

?>

它将更新一条记录，然后退出并给我:

It will update a single record, then quit and give me:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1

选择记录的第一部分工作正常，其query2不会更新所有50条记录，只是更新一条记录.也许我写错了.

The first part that selects the records works fine, its query2 that won't update all 50 records, just one. Maybe I am writing this wrong.

推荐答案

mysql_query只需要一次

mysql_query needs only one time

    $query2 = mysql_query("UPDATE contacts 
                    SET ownerID = 'Tester' 
               WHERE id = '$firstid'");

    $result2 = mysql_query($query2) or die(mysql_error());

到

    $result2 = mysql_query("UPDATE contacts 
                    SET ownerID = 'Tester' 
               WHERE id = '$firstid'");

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