Question

我需要你对我创建的php的帮助。它得到了一个警告，我不知道错误在哪里。

这是我得到的警告：

警告：mysql_query（）：用户''@ localhost'拒绝访问（使用   密码：NO）在/home/dweetcom/public_html/admin_kepuasan.php上线   10

警告：mysql_query（）：无法建立指向服务器的链接   在第10行的/home/dweetcom/public_html/admin_kepuasan.php

警告：mysql_fetch_assoc（）期望参数1是资源，   在线/home/dweetcom/public_html/admin_kepuasan.php中给出的布尔值   27

这是我的php文件：

admin_kepuasan.php

  <?php
include("connection1.php");
?>
<html>
<link rel="stylesheet" type="text/css" href='https://cdn.datatables.net/1.10.13/css/jquery.dataTables.min.css'/>

<?php

$sql="select * from borang_kepuasan";
$result=mysql_query($sql);

echo "<table id='example' class='display' cellspacing='1' width='100%'>";
echo "<thead>";
echo "<tr>";
echo "<td>Soalan 1</td>";
echo "<td>Soalan 2</td>";
echo "<td>Soalan 3</td>";
echo "<td>Soalan 4</td>";
echo "<td>Soalan 5</td>";
echo "<td>Soalan 6</td>";
echo "<td>Soalan 7</td>";
echo "<td>      Nota / Testimoni       </td>";
echo "</tr>";
echo "</thead>";
echo "<tbody>";

while ($row = mysql_fetch_assoc($result)) {
    echo "<tr>";
    echo "<td>$row[soalan1]</td>";
    echo "<td>$row[soalan2]</td>";
    echo "<td>$row[soalan3]</td>";
    echo "<td>$row[soalan4]</td>";
    echo "<td>$row[soalan5]</td>";
    echo "<td>$row[soalan6]</td>";
    echo "<td>$row[soalan7]</td>";
    echo "<td>$row[nota]</td>"; 
    echo "</tr>";
}
echo "</tbody>";
echo "</table>";
 echo "<center>";
 echo "<table>";
 echo "<tr>";
 echo "<td>";
     echo "<button style=height:30px width:80px value=Kembali onClick=location.href='admin1.php'>";
     echo "Kembali";
     echo "</button>";
 echo "</td>";
 echo "<td>";
     echo "<button style=height:30px width:80px value=Log Out onClick=location.href='logout.php'>";
     echo "Log Out";
     echo "</a>";
     echo "</button>";
echo "</center>";
?>
</html>
<script src='//code.jquery.com/jquery-1.12.4.js'></script>
<script src='https://cdn.datatables.net/1.10.13/js/jquery.dataTables.min.js'></script>
<script>
$('#example').DataTable();
</script>

connection1.php

<?php
$servername = "localhost";
$username = "dweetcom";
$password = "8weF5yaMow";

// Create connection
$conn = new mysqli($servername, $username, $password)  or die ("cannot connected");

@mysql_select_db("dweetcom_borang",$conn);
?>

Answer 1

$result=mysqli_query($conn, $sql);

在admin_kepuasan.php文件中，您需要进行以上更改。

第一个参数是连接变量（您在连接文件中创建的）

其次是查询字符串。

Answer 2

使用以下代码

$conn = new mysqli($servername, $username, $password)  or die ("cannot connected");
mysqli_query($conn,"select * from borang_kepuasan");

Answer 3

试试这个：

$con = mysqli_connect("localhost","dweetcom","8weF5yaMow","dweetcom_borang");

$query = "Enter your query";

mysqli_query($con,$query);

Answer 4

试试这段代码：＆LT; ---- ---- admin_kepuasan.php＆GT;

    <?php
require_once 'connection1.php';
?>
<html>
<link rel="stylesheet" type="text/css" href='https://cdn.datatables.net/1.10.13/css/jquery.dataTables.min.css'/>

<?php

$sql="select * from borang_kepuasan";
$result=mysqli_query($conn,$sql);

echo "<table id='example' class='display' cellspacing='1' width='100%'>";
echo "<thead>";
echo "<tr>";
echo "<td>Soalan 1</td>";
echo "<td>Soalan 2</td>";
echo "<td>Soalan 3</td>";
echo "<td>Soalan 4</td>";
echo "<td>Soalan 5</td>";
echo "<td>Soalan 6</td>";
echo "<td>Soalan 7</td>";
echo "<td>      Nota / Testimoni       </td>";
echo "</tr>";
echo "</thead>";
echo "<tbody>";

if (mysqli_num_rows($result) > 0) {
    while ($row = mysqli_fetch_assoc($result)) {
        echo "<tr>";
        echo "<td>$row[soalan1]</td>";
        echo "<td>$row[soalan2]</td>";
        echo "<td>$row[soalan3]</td>";
        echo "<td>$row[soalan4]</td>";
        echo "<td>$row[soalan5]</td>";
        echo "<td>$row[soalan6]</td>";
        echo "<td>$row[soalan7]</td>";
        echo "<td>$row[nota]</td>"; 
        echo "</tr>";
    }
} else {
    echo "0 results";
}
echo "</tbody>";
echo "</table>";
 echo "<center>";
 echo "<table>";
 echo "<tr>";
 echo "<td>";
     echo "<button style=height:30px width:80px value=Kembali onClick=location.href='admin1.php'>";
     echo "Kembali";
     echo "</button>";
 echo "</td>";
 echo "<td>";
     echo "<button style=height:30px width:80px value=Log Out onClick=location.href='logout.php'>";
     echo "Log Out";
     echo "</a>";
     echo "</button>";
echo "</center>";
?>
</html>
<script src='//code.jquery.com/jquery-1.12.4.js'></script>
<script src='https://cdn.datatables.net/1.10.13/js/jquery.dataTables.min.js'></script>
<script>
$('#example').DataTable();
</script>

这就是连接。＆LT; ---- ----- connection1.php＆GT;

<?php
$servername = "localhost";
$username = "dweetcom";
$password = "8weF5yaMow";

// Create connection
$conn = new mysqli($servername, $username, $password) ;
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

@mysqli_select_db("dweetcom_borang",$conn);
?>

也不要混用MySQL和MySQLi语法。 MySQL已从PHP 7版本中折旧。

为什么我在我的PHP中收到警告？

4 个答案: