Question

我创建了一个简单的PHP页面，它从POST请求中检索数据并将其保存在MYSQL数据库中。数据以正确的方式插入。现在，我想执行Select查询以从表中检索数据并将其显示在UI表中。 问题是Select语句出错。

这是错误：

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\abhideep_test_project\welcome.php on line 48

这是PHP代码：

<html>
<body>

<?php $name = $_POST["name"]; 
$email = $_POST["email"]; 
$mobile = $_POST["mobile"]; 
$address = $_POST["address"]; 
//=============Data Insertion=================

// Create connection
$con=mysqli_connect("localhost","root","","employee_db");

// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
/*mysqli_query($con,"INSERT INTO employee_details (employee_name, employee_email, employee_mobile, employee_address) 
VALUES ('Glenn', 'glenn@gmail.com','9830098300','2/A, Work Lane')");*/

mysqli_query($con,"INSERT INTO employee_details (employee_name, employee_email, employee_mobile, employee_address) 
VALUES ('$name','$email','$mobile','$address')");

mysqli_close($con);
//=============Data Insertion=================


//=============Data Display=================
$con=mysqli_connect("localhost","root","","employee_db");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }


$result = mysqli_query($con,"SELECT  'employee_id' ,  'employee_name' ,  'employee_email' ,  'employee_mobile' ,  'employee_address' 
FROM  'employee_details' ");

echo "<table border='1'>
<tr>
<th>Name</th>
<th>Email</th>
<th>Mobile</th>
<th>Address</th>
</tr>";

while($row = mysqli_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['employee_name'] . "</td>";
  echo "<td>" . $row['employee_email'] . "</td>";
  echo "<td>" . $row['employee_mobile'] . "</td>";
  echo "<td>" . $row['employee_address'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

mysqli_close($con);
//=============Data Display=================

?>
</body>
</html>

我哪里错了？可以做些什么来获得所需的输出？

Answer 1

您的选择查询在列和表名称周围有单引号。你应该删除它们，或者用反引号替换它们：

"SELECT  `employee_id` ,  `employee_name` ,  `employee_email` ,  `employee_mobile` ,  'employee_address' 
FROM  `employee_details` "

更新：感谢@bartfriederichs指出单引号的select子句中的引用元素无效，因为查询将返回传递的相同数据。但引用单引号的表名无效。

我的回答是假设您希望表employee_id, employee_name, employee_email, employee_mobile, employee_address中的列employee_details中的数据。

Answer 2

这不正确

$result = mysqli_query($con,"SELECT  'employee_id' ,  'employee_name' ,  'employee_email' ,  'employee_mobile' ,  'employee_address' 
FROM  'employee_details' ");

应该是

$result = mysqli_query($con,"SELECT  `employee_id` ,  `employee_name` ,  `employee_email` ,  `employee_mobile` ,  `employee_address` 
FROM  `employee_details` ");

Answer 3

您的select语句应如下所示。使用backtick作为列名

$result = mysqli_query($con,"SELECT  `employee_id` ,  `employee_name` ,  `employee_email` ,  `employee_mobile` ,  `employee_address` 
FROM  `employee_details` ");

Answer 4

嘿，你需要在你的插入变量中使用addslashes函数，

$name = addslashes($_POST["name"]); 
$email = addslashes($_POST["email"]); 
$mobile = addslashes($_POST["mobile"]); 
$address = addslashes($_POST["address"]);

然后运行您的查询，

mysqli_query($con,"INSERT INTO employee_details (employee_name, employee_email, employee_mobile, employee_address) 
VALUES ('$name','$email','$mobile','$address')");

Answer 5

<html>
<body>
<?php 
$con = mysql_connect("localhost","root","") or die("mysql error");
        $db = mysql_select_db("employee_db",$con) or die ("problem in connecting database");
        ?>
<?php

$name = $_POST["name"]; 
$email = $_POST["email"]; 
$mobile = $_POST["mobile"]; 
$address = $_POST["address"]; 


$query =mysqli_query("INSERT INTO employee_details (employee_name, employee_email, employee_mobile, employee_address) 
VALUES ('$name','$email','$mobile','$address')");

?>
<table>
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Mobile</th>
<th>Address</th>
</tr>
</thead>

<tbody>

<?php
$sql ="SELECT  'employee_id' ,  'employee_name' ,  'employee_email' ,  'employee_mobile' 'employee_address' 
FROM  'employee_details' ";
                $runquery=mysql_query($sql);
                                    $rows= mysql_num_rows($runquery);

                                    if(is_resource($runquery) && $rows>0)       
                        {
                                    while($result= mysql_fetch_array($runquery))
                            {   


                                    echo ' <tr>';
                    echo '<td>'.$result['employee_name'].'</td>';
                    echo '<td>'. $result['employee_email'].'</td>';
                    echo '<td>'.$result['employee_mobile'].'</td>';
                    echo '<td>'.$result['employee_address'].'</td></tr>';

                            }}
            ?>             
                    </tbody>
                </table>

<?php mysqli_close($con);?>
<!-- =============Data Display=================-->

</body>
</html>

MySql选择查询返回错误结果

5 个答案: