Question

   <?php

   if (isset($_COOKIE["username"])) {

$listType = $_POST['listType'];
$uid = $_COOKIE['username'];
$con = mysql_connect('localhost','xxxx','xxxx');
if(!$con)
{
        die("Could not connect: " . mysql_error());
}

$db = mysql_select_db('xxx');

$getUseridquery = mysql_query("SELECT * FROM user WHERE membername='".$uid."'") or die("error1" .mysql_error());
while ($row = mysql_fetch_array($getUseridquery)or die("error11" .mysql_error())) {
    $user_id = $row['uid'];
    typeHandler($listType, $user_id);
}

   } else {
       //something wrong
       return failure;
   }

   function typeHandler($type, $id) {
       if ($type == 'sub') {
        //get the list of subscibers if there's any
        $getSubscribe_query = mysql_query("SELECT target_id FROM subscribes WHERE owner_id='" . $id . "'") or die("error2".mysql_error());
           $subArray = mysql_fetch_array($getSubscribe_query);
           $rowcount = mysql_num_rows($getSubscribe_query) or die("error22".mysql_error());
           if ($rowcount > 0) {
            makeSubList($subArray, $id, TRUE);
           } else {
            makeSubList($subArray, $id, FALSE);
           }

       }

       if ($type == '') {

       }

   }

   function makeSubList($iArray, $id, $hasSub) {

if ($hasSub = TRUE) {
    //if user subscribed to others
    $responseArray = array();
    for ($a = 0; $a <= count($iArray); $a++) {
        makeListEntry($iArray[$a], $responseArray);
    }

    //add owner's scene
    makeListEntry($id, $responseArray);
    return $responseArray;
} else {
    //just add ower to the list
    $responseArray = array();
    makeListEntry($id, $responseArray);
    return $responseArray;
}

   }

   function makeListEntry($user_id, $responseArray) {

//scene count
$getSceneInfo = mysql_query("SELECT * FROM scene WHERE uid='" . $user_id."'") or die("error3".mysql_error());
$sceneCount = mysql_num_rows($getSubscribe_query)or die("error34".mysql_error());

//latest scene
$getLatestScene = mysql_query("SELECT * FROM scene WHERE uid='" . $user_id . "' ORDER BY time_created DESC LIMIT 1") or die("error4".mysql_error());
while($row = mysql_fetch_array($getLatestScene) or die("error44".mysql_error())){
    $title = $row['title'];
    $time = $row['time_created'];
}


//count follower
$getFollower = mysql_query("SELECT * FROM subscribes WHERE target_id='" . $user_id . "'") or die("error5".mysql_error());
$followerCount = mysql_num_rows($getFollower) or die("error54".mysql_error());

//get subscriber info
$getSubInfo = mysql_query("SELECT * FROM user WHERE uid='" . $user_id . "'") or die("error6".mysql_error());
while ($row = mysql_fetch_array($getSubInfo)or die("error66".mysql_error())){
    $dp = $row['dp_file'];
    $name = $row['name'];
}

//store data response to array
$response = array('name' => $name, 'dp' => $dp, 'title' => $title, 'uploadtime' => $time, 'scenecount' => $sceneCount, 'followercount' => $followerCount);
//store response to page
$responseArray . array_push($response);
   }

   ?>

基本上我试图从mySQL中的不同表中检索一些信息并将这些数据存储在一个数组中，这样我就可以使用AJAX将它们传递给其他JavaScript文件。

这些代码以某种方式抛出MySQL错误框，其中没有错误消息。

我尝试过分配其他字符串来识别每条错误消息，但它只显示我写的字符串，并没有解决任何错误。

P.S。我正在使用PhpMyAdmin。

请帮忙。

Answer 1

很难说这是否只是问题，但在提取时无法检查错误。如果没有剩下行，mysql_fetch_*()会返回FALSE，但不错误条件：

// Don't do this!
// If no rows are found, or as soon as you have fetched all rows,
// it will exit with a bogus error.
while($row = mysql_fetch_array($getLatestScene) or die("error44".mysql_error())){}

// Instead check for errors first, then just loop:
$getLatestScene = mysql_query("SELECT * FROM scene WHERE uid='" . $user_id . "' ORDER BY time_created DESC LIMIT 1") or die("error4".mysql_error());
while($row = mysql_fetch_array($getLatestScene)){}

注意：您的脚本容易受到SQL注入攻击。过滤输入值：

$listType = mysql_real_escape_string($_POST['listType']);
$uid = mysql_real_escape_string($_COOKIE['username']);

但是......将用户名存储在$_COOKIE中是个不错的主意。相反，您应该将该值存储在$_SESSION中。任何用户只需伪造一个cookie就可以像任何其他用户一样构成。我们无法看到您在哪里呼叫setcookie()，但您不应该这样做。而是做：

// at start of script
session_start();
// Later, store the user in $_SESSION
$_SESSION['username'] = 'the username';

Answer 2

为了您的AJAX调试目的，您的AJAX模块可以

die(JSON_encode("mysql_error",mysql_error()))

让您的调试时间更轻松。

mysql抛出空错误消息php

2 个答案: