相关疑难解决方法(0)

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)

尝试从数据库打印php代码时出错:

$query  = mysql_query('SELECT show FROM pages WHERE id = 2');
while($row = mysql_fetch_array( $query )) {
$show = $row[0];
}

错误:

警告:mysql_fetch_array()要求参数1为资源,布尔值在第18行D:\ Program Files\xampp\htdocs\rttheme9\1\admin\pages.php中给出第18行是while循环

可能重复:
mysql_fetch_array()期望参数1是资源,在select中给出boolean

以下是我的简单搜索查询,但它显示

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in 
C:\xampp\htdocs\project\p\answer.php on line 52
No Result Found 

52行:

$numserach = mysql_num_rows($search_sql);

PHP代码:

$search_sql = mysql_query("SELECT * FROM questions WHERE q_name LIKE %$search%");
            $numserach = mysql_num_rows($search_sql);

            if($numserach == 1)
            {
                echo $numserach. "Result found";
            }
            else
            {
                echo "No Result Found";
            }

谢谢:)

可能重复:
mysql_fetch_array()期望参数1是资源,在select中给出boolean

这就是我所拥有的

<?php 
            mysql_select_db("my_db", $conn);

            $result = mysql_query("SELECT name FROM card");

            while($row = mysql_fetch_array($result))
            {
            echo $row['name'];
            echo "<br />";
            }
?>

和

警告:mysql_fetch_array()要求参数1为资源,在第113行的D:\ xampp\htdocs\UFSDBSearch.php中给出布尔值

是我得到的.我不知道这意味着什么.我只是想让它回应出那张桌子上存储的卡片的名字.如果我需要告诉更多信息,请告诉我.我在这里先向您的帮助表示感谢.

我从我的数据库中取出了两个数据并将它们分开,然后将$ Kdr打印出计算出的值.

<?php
    $query = mysql_query("SELECT * FROM `high` ORDER BY `$Kdr` DESC LIMIT 20");
    while($row = mysql_fetch_array($query)){
    if ($row["kills"] != 0) {
    $Kdr = $row["kills"] / $row["deaths"];
    }       
    echo "<tr><td><a href='personalhs.php?query={$row["playerName"]}'>".$row['Runecraftlvl']."</a></td><td>".$row['kills']."</td><td>".$row['Runecraftxp']."</td><td>".$Kdr."</td></tr>";
    }
?>

忽略$ row ['Runecraftinglvl']和其他人.

基本上我希望我的表通过我在上面声明的变量"$ Kdr"来打印分割数据.

我试过这样做:

    $query = mysql_query("SELECT * FROM `high` ORDER BY `" . $kdr . "` DESC LIMIT 20");

还有这个

    $query = mysql_query("SELECT * FROM `high` ORDER BY `$Kdr` DESC LIMIT 20");

两者都给我这个错误:警告:mysql_fetch_array()期望参数1是资源,第90行/home/justxpp1/public_html/gxx/highscores.php中给出的布尔值

我究竟做错了什么？谢谢.

我收到此错误"警告:mysqli_fetch_assoc()预计参数1被mysqli_result,阵中给出"出的代码片断"searchcar.php"的

$modelmake = $_POST['model_make'];
$result = $db->select('car_information','*', 'Model_Make LIKE \'%$modelmake%\''); 
while($row = mysqli_fetch_assoc($result))
{
    echo 'Model'.$row['model_make'];
}

这是来自"database.php"的select函数的代码片段

 public function select(
        $table,
        $fields = '*',
        $where = '1=1',
        $order = '',
        $limit = '',
        $desc = false,
        $limitBegin = 0,
        $groupby = null,
        $monitoring = false
    ) //monitoring is set to true to view the actual query
    {
//  $query ='SELECT ' . $fields . ' FROM ' . $table ;
    $query = 'SELECT ' . $fields . ' FROM …

我对此查询有疑问.MySQL mysql_error在mysql_fetch_assoc方法中使用之前不会抛出.

它说

警告:mysql_fetch_assoc()期望参数1为resource,null给定

为什么PMA(PHPMyAdmin)会给我我想要的查询？(查询还可以)

码:

// last buy transactions that haven't been used yet

$transaktion = mysql_query("SELECT * FROM transaktionen 
WHERE typ = 'buy' AND used = 0 
ORDER BY id DESC") or die (mysql_error());

while ($buys = mysql_fetch_assoc($transaktionen)) {
...
}

通过我的表单提交一些东西后,我得到了欢迎部分,然后是mysql连接错误,因为mysql关闭了,当我打开它然后布尔错误时它会消失."警告:mysqli_query()要求参数1为mysqli,第25行的C:\ xampp\htdocs\welcome.php中给出布尔值"

<html>
<body>
Welcome <?php echo $_POST["name"]; ?><br>
Your email address is: <?php echo $_POST["email"]; ?><br>
Your password is <?php echo $_POST["password"]; ?><br>
You have purchased the <?php echo $_POST["sub_type"]; ?>
<?php
$mysqli_host = "localhost";
$mysql_username = "root";
$mysql_password = "123";
$site_db = "test";
$info_name = $_POST["name"];
$info_pass = $_POST["password"];
$info_email = $_POST["password"];
$sub_type = $_POST["sub_type"];

$con=mysqli_connect($mysqli_host,$mysql_username,$mysql_password,$site_db);
// Checks connection to twitch webpanel database and inserts registreation info
if (mysqli_connect_errno());
{
echo "Failed to connect to MySQL: " . …

这是我的代码......如果(mysql_num_rows($ res)> 0){我在线获取错误} {

$con= mysqli_connect("localhost","celento","password");
mysql_select_db("celento");
// Make sure we connected succesfully
if(! $con)
{
    die('Connection Failed'.mysql_error());
}
$username= $_POST['username'];
$password= $_POST['password'];


 $sql=("SELECT * FROM login WHERE username="'.$username.'" AND password="'.$password.'" LIMIT 1") or die(mysql_error());;

 $res = mysql_query($sql);

 if (mysql_num_rows($res)> 0){
             echo "Success";}
              else
             {echo"Fail";
 }
?>      

请帮我修复错误.提前致谢.

警告:mysqli_num_rows()期望参数1为mysqli_result,boolean给定iam无法登录,并且上面给出的错误被通知.建议更正.谢谢你.警告:mysqli_num_rows()期望参数1为mysqli_result,boolean给定iam无法登录,并且上面给出的错误被通知.建议更正.谢谢你.

    <html>
         <head>
             <title>Login page</title>
         </head>
             <style type='text/css'>
                 body{
                       background:url('Login.jpg');
                     }
             </style>
         <body>
             <form method ='post' action='login.php'>
                 <table width='400' border='5' align='center'>
                     <tr>
                         <td colspan='5' align='center'><h1><font color="MediumBlue">Login form</h1></font>
                         </td>

                     </tr>
                     <tr>
                         <td><font color='DarkOrange'>email</font>
                         </td>
                         <td><input type='text' name='email'/></td>
                     </tr>
                     <tr>
                         <td><font color='DarkOrange'>Password</font></td>
                         <td><input type='password' name='pass'/></td>
                     </tr>
                     <tr>
                         <td colspan='5' align='center'><input type='submit' name='login' value='login'/> </td>
                     </tr>
             </form>

                         <font color="LightSalmon"><h2><p style="position: fixed; bottom: 50%; width:100%; text-align: center"> Not registered yet?<a href='registration.php'>Sign up here</a>
                </p><h2></font>




        </body> 
    </html>
<?php
    $connection=mysqli_connect("localhost","root","","user_db");
    if(isset($_POST['login'])){
    $user_Email=$_POST['email'];
    $user_password=$_POST['pass'];

if($user_Email==''){
echo "<script>alert('please …

我有点陷入错误,在php中添加一本简单的书我在数据库中的book表中包含以下字段:

book_id book_title book_description book_category book_author date_publish qty voila ma page d'ajout d'un livre qui s'appelle save_book_query

<?php
require_once 'connect.php';
if(ISSET($_POST['save_book'])){
    $book_title = $_POST['book_title'];
    $book_desc = $_POST['book_desc'];
    $book_category = $_POST['book_category'];
    $book_author = $_POST['book_author'];
    $date_publish = $_POST['date_publish'];
    $qty = $_POST['qty'];
    $conn->query("INSERT INTO `book` (book_title,book_desc,book_category,book_author,date_publish,qty) VALUES ('', '$book_title', '$book_desc', '$book_category', '$book_author', '$date_publish', '$qty')") or die (mysqli_error() );

    echo'
        <script type = "text/javascript">
            alert("Successfully saved data");
            window.location = "book.php";
        </script>
    ';
}

出现的错误如下:

( ! ) Warning: mysqli_error() expects exactly 1 parameter, 0 …

我不知道这条线有什么问题或如何解决它，之前还好，现在我收到了这个错误：

mysqli_fetch_object() 期望参数 1 是 mysqli_result

这是我的PHP代码：

<?php
}   
if($_GET['action']=="user_info")
    {

    $userid = $_GET['user_id'];

    $query = "SELECT * FROM user WHERE user_id ='{$userid}'";
    $result = mysqli_query($link, $query);
    $user = mysqli_fetch_object($result);

    $queryt = "SELECT * FROM user_title WHERE id='".$user->title."'";
    $resultt = mysqli_query($link, $queryt);
    $rowt = mysqli_fetch_object($resultt);
    $title = $rowt->name;

        $sorgu = "select * from pub_author where user_id='$userid'";
        $publications = mysqli_query($link, $sorgu);

        while($a = mysqli_fetch_object($publications))
    {
       $ids .= $a->pub_id . ',';
    }

       $ids = rtrim($ids,",");
   $sorgu2 = "select count(id) as total , …