PHP MySQL insert query via html form, parameters not going through

Question

I tried every version of code but it is not working. I have an HTML form calling Post- PHP page to insert few fields in database. Only fields with direct text are adding, but anything coming via $_POST parameters are empty - nothing getting through no matter what I change.

HTML FORM:

<html>
<table><form method="post" action="input.php">
<tr><td>Post Title</td>
<td><input type="text" name="fdt_post_title" size="20"></td></tr>
<tr><td>Post URL</td><td><input type="text" name="fdt_post_url" size="40"></td></tr>
<tr><td>Post Description</td><td><input type="text" name="fdt_post_desc" size="40"></td></tr>
<tr><td>Post Title for FB</td><td><input type="text" name="fdt_post_title_fb" size="40"></td></tr>
<tr><td>Image URL</td><td><input type="text" name="fdt_image_url_fb" size="40"></td></tr>
<tr><td></td><td align="right"><input type="submit" name="submit" value="Submit"></td></tr>
</form></table>
</html>

input.php page:

<?php
$dt_post_title = mysql_real_escape_string($_POST['$fdt_post_title']);
$dt_post_url = mysql_real_escape_string($_POST['$fdt_post_url']);
$dt_post_desc = mysql_real_escape_string($_POST['$fdt_post_desc']);
$dt_post_title_fb = mysql_real_escape_string($_POST['$fdt_post_title_fb']);
$dt_image_url_fb = mysql_real_escape_string($_POST['$fdt_image_url_fb']);

$link = mysql_connect("localhost","root","");//database connection
mysql_select_db("bighorn1_autoshare", $link);

$now =  date();

//inserting data order
$order = "INSERT INTO topics
(id, title, url, description, facebook_post, facebook_image, facebook_pubstatus, date_published)
VALUES
(DEFAULT, '$dt_post_title', '$dt_post_url', '$dt_post_desc', '$dt_post_title_fb',     '$dt_image_url_fb', '0', '$now')";

//declare in the order variable
$result = mysql_query($order, $link);   //order executes
if($result){
echo("<br>Input data is succeed<br>");
} else {
echo("<br>Input data is fail<br>");
}
echo mysql_errno($link) . ": " . mysql_error($link) . "\n";
?>

In the table, Field id is Primary, Not Null, Auto Increment. All I am getting is id number incremented to next one in reach new row, facebook_pubstatus as 0 and date_published as0000-00-00 00:00:00`

--------------------------------------------------------------------------------------------------------------------------------
id  |   title   |   url |   description |   facebook_post   |   facebook_image  |   facebook_pubstatus  |   date_published      |
--------------------------------------------------------------------------------------------------------------------------------
1   |           |       |               |                   |                   |               0       |   0000-00-00 00:00:00 |
--------------------------------------------------------------------------------------------------------------------------------
2   |           |       |               |                   |                   |               0       |   0000-00-00 00:00:00 |
--------------------------------------------------------------------------------------------------------------------------------

and so on....

If I change to direct string in (DEFAULT, '$dt_post_title', '$dt_post_url' .... ") like (DEFAULT, 'test String 1',...... ) then this text has no issues going through.

Any magic appreciable, thanks.

Also stressing on this part too much, my brain not figuring out whats with the date now, why all zeros.

Thanks

Please, don't use mysql_* functions, They are no longer maintained and are officially deprecated. Learn about prepared statements instead, and use PDO or MySQLi. This article will help you decide. — Jay Blanchard
– Jay Blanchard, Commented Nov 6, 2014 at 20:10
Add error reporting to the top of your file(s) right after your opening <?php tag error_reporting(E_ALL); ini_set('display_errors', 1); — Jay Blanchard
– Jay Blanchard, Commented Nov 6, 2014 at 20:12
Don't just "look into it", make the changes or one day you will be on the receiving end of an example of what happens when you continue to use mysql_* functions. I'd hate to see you have to recode everything because your PHP version was upgraded and your sites stopped working. — Jay Blanchard
– Jay Blanchard, Commented Nov 6, 2014 at 20:15
@JayBlanchard Yup I will definitely change to mysqli, new to php (beginner - long way to go yet) Thanks. For this question, I danmullen answer kinda worked out. But yea I will take your advice and now will change the code to mysqli api. — user2350586
– user2350586, Commented Nov 6, 2014 at 20:21

danmullen · Accepted Answer · 2014-11-06 20:11:14Z

3

You have an extra '$' sign in your variables. Change to this:

$dt_post_title = mysql_real_escape_string($_POST['fdt_post_title']);
$dt_post_url = mysql_real_escape_string($_POST['fdt_post_url']);
$dt_post_desc = mysql_real_escape_string($_POST['fdt_post_desc']);
$dt_post_title_fb = mysql_real_escape_string($_POST['fdt_post_title_fb']);
$dt_image_url_fb = mysql_real_escape_string($_POST['fdt_image_url_fb']);

answered Nov 6, 2014 at 20:11

danmullen

2,5003 gold badges22 silver badges28 bronze badges

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

1 Comment

user2350586 Over a year ago

Thanks @danmullen . Stupid question I think I asked :) happens sometimes when working for long hours. Beautifully done.

Community · Accepted Answer · 2020-06-20 09:12:55Z

You need to pass the link_identifier to your mysql_real_escape_string. http://php.net/manual/en/function.mysql-real-escape-string.php

The MySQL connection. If the link identifier is not specified, the last link opened by mysql_connect() is assumed. If no such link is found, it will try to create one as if mysql_connect() was called with no arguments. If no connection is found or established, an E_WARNING level error is generated.

Returns the escaped string, or FALSE on error.

You are probably getting the value "false" in each of your variables right now. Try it this way:

$link = mysql_connect("localhost","root","");//database connection
$dt_post_title = mysql_real_escape_string($_POST['fdt_post_title'], $link);
$dt_post_url = mysql_real_escape_string($_POST['fdt_post_url'], $link);
$dt_post_desc = mysql_real_escape_string($_POST['fdt_post_desc'], $link);
$dt_post_title_fb = mysql_real_escape_string($_POST['fdt_post_title_fb'], $link);
$dt_image_url_fb = mysql_real_escape_string($_POST['fdt_image_url_fb'], $link);

Collectives™ on Stack Overflow

PHP MySQL insert query via html form, parameters not going through

2 Answers 2

1 Comment

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related