Question

我目前遇到的问题是我的查询返回成功但是在检查我的SQL数据库时没有发生实际更新。奇怪的是，当我将相同的查询复制到phpMyAdmin时，成功返回响应并且查询工作正常，行更新。（注意：我很清楚SQL注入的高风险，但mysqli_escape_string因某些原因无法正常工作，所以当我进入该网站时，我会担心这一点生产阶段。）

的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会返回如下所示的值：'abcd123','dcba321','hello123'，这会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工作得很好，但是当通过php执行时，响应会返回成功但是没有实际更新的行，所以我不确定发生了什么我的php error.log就像哨子一样干净。

另外，如果您想知道我的require_once connection.php 文件是什么样的，这会将我连接到数据库，它是以下内容：

$link = ($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");
}

Answer 1

愚蠢的我，我不确定它为什么会返回成功的查询，但是将列ID an_id包装在单引号中是个问题

查询返回成功但没有更新行

1 个答案: