[英]Query Returns Success But no Rows Updated
我目前遇到查询返回成功的问题,但是当我检查SQL数据库时没有发生实际更新。 奇怪的是,当我将相同的确切查询复制到phpMyAdmin中时,成功返回了响应,并且查询工作正常,行已更新。 ( 注意:我很清楚SQL注入的高风险,但是由于某些原因mysqli_escape_string
无法正常工作,因此在进入生产阶段时我会为此担心。 )
script.php
$fave = json_decode($_POST['af']);
$unfave = json_decode($_POST['uf']);
$fave = "'".implode("','", $fave)."'";
$unfave = "'".implode("','", $unfave)."'";
if ($fave !== "''"){
$fq = "UPDATE post SET fave='1' WHERE 'an_id' IN ($fave) AND bid='$bizusr' AND fave='0'";
$r_fq = mysqli_query($GLOBALS["___mysqli_ston"], $fq);
$ar_fq = mysqli_affected_rows($GLOBALS["___mysqli_ston"]);
} else {
$r_fq = 1;
$ar_fq = 0;
}
if ($unfave !== "''"){
$ufq = "UPDATE post SET fave='0' WHERE 'an_id' IN ($unfave) AND bid='$bizusr' AND fave='1'";
$r_ufq = mysqli_query($GLOBALS["___mysqli_ston"], $ufq);
$ar_ufq = mysqli_affected_rows($GLOBALS["___mysqli_ston"]);
} else {
$r_ufq = 1;
$ar_ufq = 0;
}
if ($r_fq && $r_ufq){
$output = json_encode(array('type'=>'error', 'text' => "Favourites have been updated successfully. You've added $ar_fq favorites and removed $ar_ufq favorites." ));
die($output);
}
if (!$r_fq && $r_ufq){
$output = json_encode(array('type'=>'error', 'text' => "We've successfully favorited $ar_fq links, however there was an issue in unfavoriting some links, try refreshing." ));
die($output);
}
if ($r_fq && !$r_ufq){
$output = json_encode(array('type'=>'error', 'text' => "We've successfully unfavorited $ar_ufq links, however there was an issue in favoriting some links, try refreshing." ));
die($output);
}
if (!$r_fq && !$r_ufq){
$output = json_encode(array('type'=>'error', 'text' => "There was an error in updating your favorited links." ));
die($output);
}
// $un = mysqli_prepare($GLOBALS["___mysqli_ston"], "UPDATE analytics SET fave='0' WHERE an_id IN (?) AND bid= ? AND fave='1'");
// $fa = mysqli_prepare($GLOBALS["___mysqli_ston"], "UPDATE analytics SET fave='1' WHERE an_id IN (?) AND bid= ? AND fave='0'");
// mysqli_stmt_bind_param($un, 'ss', $unfave, $blockject);
// $a = mysqli_stmt_execute($un);
// mysqli_stmt_close($un);
// mysqli_stmt_bind_param($fa, 'ss', $fave, $blockject);
// $b = mysqli_stmt_execute($fa);
// mysqli_stmt_close($fa);
变量$fave
和$unfave
将返回如下值: 'abcd123','dcba321','hello123'
,这将使查询看起来像这样:
UPDATE post SET fave='0' WHERE 'an_id' IN ('abcd123','dcba321','hello123') AND bid='$bizusr' AND fave='1';
现在,将查询输入phpMyAdmin
可以正常工作,但是通过php进行查询时,响应返回成功,但是实际上没有行被更新,因此我不确定发生了什么,因为我的php error.log是否干净吹口哨。
另外,如果您想知道将我连接到数据库的我的require_once
connection.php文件是什么样子,请执行以下操作:
$link = ($GLOBALS["___mysqli_ston"] = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD));
if(!$link) {
die('Failed to connect to server: ' . ((is_object($GLOBALS["___mysqli_ston"])) ? mysqli_error($GLOBALS["___mysqli_ston"]) : (($___mysqli_res = mysqli_connect_error()) ? $___mysqli_res : false)));
}
//Select database
$db = ((bool)mysqli_query($GLOBALS["___mysqli_ston"], "USE " . constant('DB_DATABASE')));
if(!$db) {
die("Unable to select database");
}
愚蠢的我,我不确定为什么它会返回成功的查询,但是问题是将列ID an_id
在单引号中
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