PHP code executes but no output

Question

I am trying to learn "searching elements from mysql database using php".

For this I created a database named randomdata. In randomdata database there is a table named randomtable. In this table there are four columns: Name, Surname, Email and Gender.

I want to search people by there Gender. For this I tried following query.

 $query="SELECT * FROM randomtable WHERE Gender =' ".$gender . " ' ";

I tried both, GET and POST functions. But still I am not able to take output. I am using these.

Windows 8
Wampserver

Notepad++

I restarted server and PC, but nothing changed. Below is my complete code.

Find Entries:
Male
Female

<?php
if(isset($_POST['submit']))
{
     echo $gender=$_POST['$gender'];
    $connect=mysql_connect("127.0.0.1","root","", "randomdata");
    if($connect)
    {
    //echo 'I am connected';
    $query="SELECT * FROM randomtable WHERE Gender =' ".$gender . " ' ";
        echo  $query;
         $results=mysqli_query( $connect,$query);
        while($row = mysqli_fetch_array($results))
        {
            echo $row['Name']  . "<br/>" . $row['Surname']  . "<br/>" . $row['Email']  .   "<br/>" ;
        }
    }
    else
    {
        die(mysql_error());
    }
}
?>

print your query and run it in phpmyadmin.than check is all things goes well ? — Harshal
– Harshal, Commented Jan 7, 2014 at 13:57
Maybe changing $_POST['$gender'] to $_POST['gender']? Shooting in the dark. Maybe your $gender param is empty. — divaka
– divaka, Commented Jan 7, 2014 at 13:59
@user3168766 yeah you are putting $gender inside the POST['$gender'] gender you getting form through POST method is not a php variable — Waqas
– Waqas, Commented Jan 7, 2014 at 14:47
@Rana Muhammad Waqas I solved my problem, thank you for your co-operation. — Umar Farooq
– Umar Farooq, Commented Jan 7, 2014 at 14:52
@Rana Muhammad Waqas Answer having green color sign, was very helpful for me. — Umar Farooq
– Umar Farooq, Commented Jan 7, 2014 at 14:53

The Humble Rat · Accepted Answer · 2014-01-07 14:29:17Z

3

It looks like the issue is with assigning the POST variable to $gender

Currently you are using

echo $gender=$_POST['$gender'];

Please try changing this to

$gender=$_POST['gender'];

UPDATE

After testing your code it seems the isset is the issue. There is never a POST['Submit'].

To fix this you need the name attribute in the submit input ie

<input type="submit" Value="Search" name="Submit"/>

Also in the query you have spaces either side of the $gender variable. I now have the code working, try with this.

<html>
<body>
 Find Entries: <br> 
 <form action="" method="POST"> 
 <input type="radio" name="gender" Value="Male"> Male </input>
 <br> 
 <input type="radio" name="gender" Value="Female"> Female </input>
 <br> 
 <input type="submit" Value="Search"/> 
</form>

<?php
if(isset($_POST['gender']))
{
     //print_r($_POST);
     $gender=$_POST['gender'];
     $connect=mysqli_connect("127.0.0.1","root","password", "randomdata");
     if($connect)
     {
        //echo 'I am connected';
        $query="SELECT * FROM randomtable WHERE Gender = '".$gender . "' ";
        //echo  $query;
        $results=mysqli_query( $connect,$query);
        while($row = mysqli_fetch_array($results))
        {
            echo $row['Name']  . "<br/>" . $row['Surname']  . "<br/>" . $row['Email']  .   "<br/>" ;
        }
     }
     else
     {
        die(mysql_error());
     }
}
?>

edited Jan 7, 2014 at 14:29

answered Jan 7, 2014 at 14:00

The Humble Rat

4,7468 gold badges44 silver badges74 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

Umar Farooq Over a year ago

I can send you whole code. If you want to see. It is small code consisting of 34 lines.

The Humble Rat Over a year ago

@user3168766 Please provide the whole code in your question and also, please provide a print of the POST data, so we can ensure everything is as expected.

The Humble Rat Over a year ago

@user3168766 Also please clarify how you save the gender in the table, ie M/F or Male/Female. This may be the issue.

Umar Farooq Over a year ago

I saved Male and Female.

Umar Farooq Over a year ago

<html> <body> Find Entries: <br> <form action="" method=" POST"> <input type="radio" name="gender" Value="Male"> Male </input><br> <input type="radio" name="gender" Value="Female"> Female </input><br> <input type="submit" Value="Search"/> </form> </body > </html>

|

Community · Accepted Answer · 2023-11-17 20:33:26Z

3

mysql_connect should be mysqli_connect

edited Nov 17, 2023 at 20:33

CommunityBot

11 silver badge

answered Jan 7, 2014 at 13:57

Tomás

3,6063 gold badges24 silver badges39 bronze badges

Comments

user1844933 · Accepted Answer · 2014-01-07 13:54:27Z

1

Try this...

$query="SELECT * FROM randomtable WHERE Gender ='".$gender . "'";

Removed extra spaces in query

answered Jan 7, 2014 at 13:54

user1844933

3,4132 gold badges28 silver badges44 bronze badges

3 Comments

Umar Farooq Over a year ago

I copied this excerpt and replace my previous query with it. Nothing changed.

user1844933 Over a year ago

as per @The Humble Rat change echo $gender=$_POST['$gender']; to $gender=$_POST['gender'];

Umar Farooq Over a year ago

@tomas I tried your suggestion. I am sorry to say nothing changed. I restarted server too.

tomsowerby · Accepted Answer · 2014-01-07 14:06:12Z

0

Is there no output at all? Very odd indeed.

What are you posting?

Are you sure that you're including 'submit' in your test post?

If this doesn't help, perhaps there's a more severe error that's not allowing the script to run? Are you able to see the php or apache error logs?

answered Jan 7, 2014 at 14:06

tomsowerby

4981 gold badge3 silver badges10 bronze badges

5 Comments

Umar Farooq Over a year ago

Yes I am including submit.

Umar Farooq Over a year ago

if(isset($_POST['submit'])) you can see this.

tomsowerby Over a year ago

Just checking, sometimes it's easy to miss. If it wasn't present then it would explain why nothing else was running.

Chandrayya G K Over a year ago

Add the comments below the question if you need clarification.

tomsowerby Over a year ago

@chandrayya-g-k I tried, I was told that my rep wasn't high enough yet. I'll work harder....

Yami · Accepted Answer · 2014-01-07 14:11:19Z

0

I guess you were intending to assign the "submit" index to the $gender variable instead of the "gender" index? Try the code below:

<?php
if(isset($_POST['submit']))
{
    $gender = $_POST['submit'];
    $connect = mysql_connect("127.0.0.1", "root", "", "randomdata");
    if($connect)
    {
    $query = 'SELECT * FROM randomtable WHERE Gender = "' .$gender .'"';
        $results = mysqli_query($connect, $query);
        while($row = mysqli_fetch_array($results))
        {
            echo $row['Name']  . "<br/>" . $row['Surname']  . "<br/>" . $row['Email']  .   "<br/>" ;
        }
    }
    else
    {
        die(mysql_error());
    }
}
?>

answered Jan 7, 2014 at 14:11

Yami

1,4253 gold badges11 silver badges16 bronze badges

1 Comment

Umar Farooq Over a year ago

I didn't received any output.

Collectives™ on Stack Overflow

PHP code executes but no output

5 Answers 5

8 Comments

Comments

3 Comments

5 Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

8 Comments

Comments

3 Comments

5 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related