CODEWITHSUNDEEP
phpmysql
user3840545
user3840545

Reputation: 25

how to handle selection form with input text in php mysql to be saved

Hello I'm trying to save the value in the text input field to be saved in database.

What I'm trying to do is If I choose singing the singing will be saved and same with dancing If I choose Other in the selection the input will show and I'll be entering the value of other and that value I entered in the input text will be saved in the database.

The problem with my code is If I choose other and put a value in the text, the data value will be saved is only other.

form

Talent:
    <form method=post action=test2.php>
    <select name="talent" onchange="if( this.value=='other' )   { this.form['other'].style.visibility='visible' }else     { this.form['other'].style.visibility='hidden' };" required/>
    <option value="Dancing">Dancing</option>
    <option value="Singing">Singing</option>
    <option value="other">Other</option>
    <input type="text" name="other" style="visibility:hidden;" />
    </select>
    <button type=submit>
</form>

Test2.php

$sql="INSERT INTO database (Talent) VALUES('$_POST[Talent]',)";

Upvotes: 0

Views: 989

Answers (2)

rack_nilesh
rack_nilesh

Reputation: 553

You are inserting $_POST[Talent] in database. So if you select 'Other' option in dropdown,then instead of text box value,'Other' value will be inserted in db.

Try this

if($_POST['Talent'] == 'other'])
{
    $other = $_POST['other'];//considering 'other' as name of your rextbox
    $sql="INSERT INTO database (Talent) VALUES('$other',)";
}
else
{
$sql="INSERT INTO database (Talent) VALUES('$_POST['Talent']',)";
}

Upvotes: 0

Mainz007
Mainz007

Reputation: 543

You're input field saves the value in $_POST['other']. So you need to check, if other is selected and then insert the input field. So you're final solution will be:

if($_POST['talent'] == "other"])
    $talent = $_POST['other'];
else
    $talent = $_POST['talent'];

$sql="INSERT INTO database (Talent) VALUES('$talent',)";

Hint Never directly insert POST-values in your database. It's a big risk. For further informations read here: http://www.codeproject.com/Tips/407379/How-to-Protect-from-SQL-Injection-in-PhP-based-web Still you also need to escape it. Just read what you can do ;)

Upvotes: 1

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