+ 4
Form Submit help!
I made a video of the problem you can watch it here: http://metrikcorp.com/test/rate/video/ so i have a form and every time some one goes to it automatically submits the form blank. it will also do this if you refresh the page. How can i go about stopping this? see the form live here: http://www.metrikcorp.com/test/rate <!DOCTYPE html> <html> <head> <title>Image Rating</title> <script> function validateForm() { var x = document.forms["Rating"]["rating_number"].value; if (x
10 Answers
+ 5
when ever you go to my form it will automatically submit into mysql database (as shown in video) I do not want it to submit unless the submit button is pressed. does that make more sense?
+ 4
Sandeep chatterjee: i am still seeing the issue on my end. every time i go to it or hit the refresh it adds one vote to my database and it is blank.
I made a video of the problem you can watch it here: http://metrikcorp.com/test/rate/video/
+ 3
What do you want actually, print error message if text field is empty, right?
+ 3
Aditya kumar pandey:
ya thats fine. as long as it stops the form from submiting blank info when load or refreshed.
+ 3
what? I did not understand it.
+ 3
Chibu Nathy: what do you mean? when you click on the video link nothing happens? if so what kinda browser(google, IE, firefox, ect) you using and what kinda device? desktop, mobile, tablet?
+ 3
Than this problem is arise due to your SQL implementation. Please show that code. I know you want to find answer, but hey How can I help if I don't know the code. Sorry for late reply
+ 3
Aditya kumar pandey: sorry for the long delay in response. Here is my code.
<?php
$con = mysql_connect("localhost","UserName","Password");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("rateImage", $con);
$sql="INSERT INTO post_rating (rating_number)
VALUES
('$_POST[rating_number]')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
?>
<?php
$Get = "select sum(rating_number) from post_rating";
$q = mysql_query($Get);
$row = mysql_fetch_array($q);
$NUMB = $row[0];
?>
<?php
$link = mysql_connect("localhost", "UserName", "Password");
mysql_select_db("rateImage", $link);
$result = mysql_query("SELECT * FROM post_rating", $link);
$num_rows = mysql_num_rows($result);
$NumRow = "$num_rows";
?>
+ 2
Sorry, I was forget about that. What is your PHP version of your server. If it is 5.5 or higher please use mysqli function and add i after MySQL in all function. Also, insert data only if $_POST[rating_number] is not null. Please forgive me. Sorry 😭😭😭
+ 1
when you click on the play video image nothing will show its blank.i thought it should be image only or img ??
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