MySql Select query returning erroneous results

Question

I have created a simple PHP page which retrieves data from a POST request and saves it in MYSQL database. The data is getting inserted in the proper way. Now, I want to execute a Select query to retrieve the data from the table and show it in a UI Table. The problem is that the Select statement is giving an error.

This is the error:

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

This is the PHP code:

<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', '[email protected]','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>

Where am I going wrong? What can be done to get the desired output?

vee · Accepted Answer · 2014-01-29 08:05:33Z

2

Your select query has single quotes around columns and table name. You should remove them, or replace them with a backtick:

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

Update: Thanks to @bartfriederichs for pointing out that quoting elements in select clause with single quote is not invalid as the query would return the same data passed. Quoting table name with single quote however is invalid.

My answer assumes that you want data in columns employee_id, employee_name, employee_email, employee_mobile, employee_address from table employee_details.

edited Jan 29, 2014 at 8:05

answered Jan 29, 2014 at 7:49

vee

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2 Comments

Bart Friederichs Over a year ago

Around the columns is not invalid, it will just return the value. Around the table is.

vee Over a year ago

@BartFriederichs, yes thank you for catching the sentence. Updated answer removing incorrect statement.

Abhik Chakraborty · Accepted Answer · 2014-01-29 07:49:35Z

2

This is not correct

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

should be

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

answered Jan 29, 2014 at 7:49

Abhik Chakraborty

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Comments

Nagaraj S · Accepted Answer · 2014-01-29 07:52:13Z

0

Your select statement should be like this..Use backtick for column name

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

answered Jan 29, 2014 at 7:52

Nagaraj S

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Comments

Shivam · Accepted Answer · 2014-01-29 08:16:24Z

0

Hey you need to use addslashes function in your insert variables like this,

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

then run your query,

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

answered Jan 29, 2014 at 8:16

Shivam

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2 Comments

Abhik Chakraborty Over a year ago

Bad idea of adding addslashes().. Why not use mysqli_real_escape_string()

Shivam Over a year ago

Hey it's not the bad idea. We should handle these type of things in server side not in the DB.

Rahul rajoria · Accepted Answer · 2014-01-29 08:31:04Z

<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>

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