(PDO)UPDATE不更新与WHERE变量
问题描述:
我的问题是,只要我有与GuildName在哪里它不起作用。它根本不会更新任何东西。
只要我有id = 1(有100个公会,所以设置id = 1不是一个选项),它确实有效。(PDO)UPDATE不更新与WHERE变量
$form = $_POST;
$boss = $form['bossname'];
$gname = $form['guildname'];
$screen = $form['screenshot'];
$log = $form['logs'];
$defeat = $form['defeat'];
if(isset($_POST['edit-guild'])){
$Statement = $conn->prepare("UPDATE $boss SET `Bossname` =:boss, `GuildName` =:gname, `Screenshot` =:screen, `Link` =:link, `KillTime` =:defeattime, `KillYN` =:kill WHERE `GuildName`=:gname");
// EXECUTING ARRAY FOR ^GUILD INFORMATION
$Statement->execute(array(
"boss" => $boss,
"gname" => $gname,
"screen" => $screen,
"link" => $log,
"defeattime" => $defeat,
"kill" => 'Yes'
));
}
只要我有WHERE id = 1它确实工作。
$Statement = $conn->prepare("UPDATE $boss SET `Bossname` =:boss, `GuildName` =:gname, `Screenshot` =:screen, `Link` =:link, `KillTime` =:defeattime, `KillYN` =:kill WHERE id=1");
答
如上所述,你不能重用相同的参数。尝试给它一个不同的名称:
if(isset($_POST['edit-guild'])){
$Statement = $conn->prepare("UPDATE $boss SET `Bossname` =:boss, `GuildName` =:gname, `Screenshot` =:screen, `Link` =:link, `KillTime` =:defeattime, `KillYN` =:kill WHERE `GuildName`=:gname2");
// EXECUTING ARRAY FOR ^GUILD INFORMATION
$Statement->execute(array(
"boss" => $boss,
"gname" => $gname,
"gname2" => $gname,
"screen" => $screen,
"link" => $log,
"defeattime" => $defeat,
"kill" => 'Yes'
));
}
答
从更新集中删除gname
字段。
$Statement = $conn->prepare("UPDATE $boss SET `Bossname` =:boss, `Screenshot` =:screen, `Link` =:link, `KillTime` =:defeattime, `KillYN` =:kill WHERE `GuildName`=:gname");
,如果你认为gname
处于更新SET
,你可以尝试其他不同的名称或试图改变where条件像id
另一场不能重复使用相同的参数名称非常重要的。 –
为什么用相同的值更新'gname'? –
这是正确的'更新$老板SET' $老板作为参数和tabelname? – JustOnUnderMillions