我想要一个我正在查看的论坛ID的所有主题的列表,但我也希望获得有关我正在查看的论坛的信息,例如名称和描述.我的查询不起作用:
警告:mysql_num_rows()期望参数1是资源,第11行的forum.php中给出了boolean
如何在此查询中加入我的论坛表？
$tresult = mysql_query("SELECT * FROM threads WHERE threads.forumID = ".intval($_GET['forumID'])." LEFT JOIN forum ON forum.id = threads.forumID");
// does the forum even exist?
if (mysql_num_rows($tresult) < 1) {
// Show error and return
echo "The forum you are looking for appears to be missing.";
return false;
}
if ($is_l
我的桌子:
forum:id,名称,描述threads:id,forumID,标题,正文,日期这是我的代码:我认为没有什么可以导致这个......想法？
警告:mysql_fetch_array()期望参数1是资源,给定布尔值.
<?php
include_once "mysql_connect.php";
if ($_POST['parse_var'] == "contactform"){
if(is_array($categories)) $whereCond = "in '".implode(",",$categories); else
$wherecond = "= ".$categories;
$dropdownValue = $_POST['dropdown'];
$dropdownValue = mysql_real_escape_string($dropdownValue);
$dropdownValue = eregi_replace("`", "", $dropdownValue);
$searchField= $_POST['searchinput'];
$searchField = mysql_real_escape_string($searchField);
$searchField = eregi_replace("`", "", $searchField);
if ($dropdownValue == "phone"){
$sql = mysql_query("SELECT * FROM pcparts WHERE phone='$searchField'");
while($row1 = mysql_fetch_array($sql)){
$arrayuserinfo[] = array(
'phone' => $row1["phone"],
'name' => $row1["name"],
'city' => $row1["city"],
'state' => $row1["state"],
'address' => $row1["address"],
'zip' => $row1["zip"],
);
}; …我在循环访问SQL查询时遇到问题,这是我的代码,它从查询结果中读取每个结果.
// Check if PCID is already inside DB > ActivePCS
$CHECK_PCID_Q = mysql_query("SELECT * FROM ActivePCS where CustomerID='$CIDX' AND Customer_Email='$EMAIL' AND Pc_Name='$PCID'");
//$CHECK_PCID_R = mysql_fetch_array($CHECK_PCID_Q);
$isfound = 0;
while($CHECK_PCID_R = mysql_fetch_array($CHECK_PCID_Q)) // <<-- error is here
{
if($CHECK_PCID_R['PC_Name'] == $PCID)
{
$isfound = 1;
break;
}
}
但我得到了这个错误:
warning mysql_fetch_array() expects parameter 1 to be resource boolean given
在我的localhost它工作正常,但当我上传到我的服务器时,它给出了这个错误.
编辑:对不起,我的错,我的查询有错误的表名.....
好的,所以我知道'为什么我用mysql_fetch_array得到这个警告...'的问题已被问过几次,我的问题是所有接受的答案都说服务器吐出这个警告的原因是因为查询本身是不正确的...这不是我的情况.
以下是代码:
$dbhost = "host";
$dbuser = "user";
$dbpass = "pass";
$dbname= "db";
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($dbname) or die ('<META HTTP-EQUIV="Refresh" CONTENT="0;URL=Failed.php?dberror=1">');
$token = mysql_escape_string($_GET['token']);
$query = "SELECT * FROM newuser WHERE token='$token'";
$result = mysql_query($query) or die(mysql_error());
while($row=mysql_fetch_array($result)) {
do stuff...
}
'while'语句中的所有内容都正常执行 - 它会对DB进行一些更改,我可以对其进行验证.更重要的是,查询永远不会泄漏任何错误详细信息.我已经尝试测试$ result === false并询问错误信息的情况,但它不会返回任何内容.据我所知,查询字符串很好,并没有失败.
我在这做错了什么？有没有其他原因为什么PHP不喜欢我的参数除了SQL语句之外的参数是坏的(再次,我相信它不坏)？
另外,我知道我应该使用mysqli/PDO ....我计划在不久的将来切换到那个,但是我只是想把这个工作拉出去,我不知道它为什么会赢得'吨.在这一点上,这更像是个人的事情......
感谢您的帮助,如果您需要任何其他信息,请与我们联系.(PHP版本5.3)
这是查询字符串($ query)的回显:
SELECT * FROM newuser WHERE token='6e194f2db1223015f404e92d35265a1b'
这里是查询结果的var_dump($ result):类型的资源(3)(mysql结果)
您好，我正在尝试将 mysql 更改为 mysqli，但我的脚本无法正常工作。更改为 mysqli 后
数据库.php
<?php
include("constants.php");
class MySQLDB
{
var $connection; //The MySQL database connection
var $num_active_users; //Number of active users viewing site
var $num_active_guests; //Number of active guests viewing site
var $num_members; //Number of signed-up users
/* Note: call getNumMembers() to access $num_members! */
/* Class constructor */
function MySQLDB(){
/* Make connection to database */
$this->connection= mysqli_connect(DB_SERVER, DB_USER, DB_PASS,DB_NAME);
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
} …一切都很好,突然发生这种情况:
PHP警告:mysql_fetch_assoc()期望参数1是资源,布尔值在第17行的G:\ PleskVhosts\modelcharm.com\httpdocs\core\functions\users.php中给出
function user_data($user_id) {
$data = array();
$user_id = (int)$user_id;
$func_num_args = func_num_args();
$func_get_args = func_get_args();
if ($func_num_args > 1) {
unset($func_get_args[0]);
$fields = '`' . implode('`, `', $func_get_args) . '`';
$data = mysql_fetch_assoc(mysql_query("SELECT $fields FROM `users` WHERE `user_id` = $user_id"));
return $data;
}}
代码中有什么问题吗？
你好,你可以查看我的代码.我这里有三个未定义的变量,我真的不知道要修复什么
- 注意:未定义索引:第70行中的用户名
- 注意:未定义的索引:第71行的密码
- 警告:mysql_fetch_array()期望参数1是资源,第86行给出布尔值
<!DOCTYPE HTML>
<?php
//Start session
session_start();
//Unset the variables stored in session
unset($_SESSION['SESS_Username']);
unset($_SESSION['SESS_Password']);
?>
<html>
<head>
<title> Ilokandroid Dynamic Web </title>
<meta name="author" content="jessiemaymasaoay and jaysonpinzon" >
<meta name="description" content="Student MiniSystem" >
<meta name="keywords" content="HTML,CSS,XML,Javascript, miniSystem Web Page" />
<style type="text/css" media="screen">
body{background: url(assets/images/back1.png) top no-repeat fixed;}
.upper
{
width:800px;height:100px; margin:auto; border: 5px; padding:2px;
}
#form
{
width:400px;height:190px; margin:auto; border: 2px solid black; padding:10px;
text-align: justify; margin-top: 150px; background:#008080 ; font-size: 1p6x;
}
</style>
</script> …我有以下代码有两个查询.第一个获取将在下一个查询中使用的ID.我在第二个查询上出现此错误:
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\b\pprincipal.php on line 132
这是第一个查询(对于该表中存在数据的用户的ID)
$query2 = mysql_query("SELECT DISTINCT competencias, id_user FROM ce");
$ut = array();
while ($result = mysql_fetch_array($query2)) $ut[] = $result;
foreach ($ut as $u){
$id_user=$u[1];
echo "<li class='has-sub'><a>" .$u[0]. "</a>";
}
这是查询给我的错误:
$query3 = mysql_query("SELECT id, nome FROM utilizadores WHERE id=$id_user");
$utt = array();
while ($result2 = mysql_fetch_array($query3, $con)) $utt[] = $result2;
foreach ($utt as $u2){
echo "<ul><li><a>". $u2[0]. "</a></li>";
}
第二个查询是正确的,因为我已经回应了数组,它没关系.但它给了我WHILE循环错误.
这是我的私人服务器realmeye的代码http://www.thenoxusproject.org/realmeye/index.php 我一直收到这个错误,我不知道如何解决它,请帮忙吗？
致命错误:在第112行的/home/noxus/public_html/realmeye/index.php中调用非对象的成员函数query()
================================================== =============
index.php代码如下
<?php
require_once('template/config.php');
// database info
define('HOST','MY VPS IP');
define('DB','rotmg');
define('USER','root');
define('PASS','DB PASSWORD');
function connect_db()
{
$pdo = new PDO('mysql:host='.HOST.';dbname='.DB,USER,PASS);
return $db;
}
?>
<html lang="en">
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1, target-densityDpi=device-dpi">
<title>Player List</title>
<link rel="shortcut icon" href="img/favicon.ico">
<meta name="msapplication-config" content="none">
<link rel="canonical" href="/top-players-by-fame">
<meta name="keywords" content="realm of the mad god, rotmg, statistics, stats,top players,by fame">
<meta name="description" content="Top Players in Realm of the Mad God by fame 1-101.">
<link href="css/bootstrap.min.css" rel="stylesheet"> …致命错误：调用成员函数 execute() on boolean in ---- 第 18 行
我不断收到此错误。我想要发生的是当您单击帖子上的删除按钮时，它将删除它。这是第 18 行
$result->execute(array(':id' => $_GET['delpost']));
PHP
if (isset($_GET['delpost'])) {
$result = mysqli_query( $connection,'DELETE FROM blog WHERE id = :id') ;
$result->execute(array(':id' => $_GET['delpost']));
header('Location: blog.php?action=deleted');
exit;
}