Accessing a variable from another file in PHP, SQL

Question

<?php
$paseoLocationTo=$_POST['locationTo'];
$paseoLocationFrom=$_POST['locationFrom'];
$PTime=$_POST['time'];

echo"The value of Location to is $paseoLocationTo </br>";
echo"The value of Location from is $paseoLocationFrom </br>";
echo"The value of time is $PTime </br>";

mysql_connect('localhost', 'root', '')
            or die(mysql_error());

    mysql_select_db('shuttle_service_system') 
    or die(mysql_error());

$TripID =mysql_query("
    SELECT DISTINCT Trip_ID as 'TripID'
    FROM trip
    WHERE Timeslot LIKE '$PTime' AND Location_From Like '$paseoLocationFrom' AND Location_To    LIKE '$paseoLocationTo'     
");
    echo "<form action='LastPage.php' method='post'>";
    while($check = mysql_fetch_array($TripID))
    **echo "<name='TripID' id='TripID'>" . $check['TripID'] . " ";**
    echo "<p class='sure'> Are you sure with your reservation? </p>";
    echo"<input type='submit' value='Submit' class='Log'>";
    echo"</form";
?>

From another php file, this the LastPage.php

<?php
**$TripID=$_POST['TripID'];
echo"The value of trip ID is $TripID </br>";**
?>

Hi guys I was wondering why I can't access the "TripID" variable in the other php file? I was accessing it before but now there seems to be a problem, am I doing it right? I'm sorry a php and SQL newbie.

This is by far the most invalid HTML and worst PHP I've ever seen. encourage developers to keep their code readable, valid and cross-browser compatible — Daniel W.
– Daniel W., Commented Nov 13, 2013 at 14:17
Its a great think if this works without showing any error.. :) — Vinoth Babu
– Vinoth Babu, Commented Nov 13, 2013 at 14:17

D4V1D · Accepted Answer · 2013-11-13 14:13:02Z

1

You need to add

 <input type="text" name="TripID" value="'.$check['TripID'].'" ... />

in your form in order to retrieve values with $_POST['TripID'].

There is no such thing as

 **echo "<name='TripID' id='TripID'>" . $check['TripID'] . " ";**

which was found in your code.

answered Nov 13, 2013 at 14:13

D4V1D

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1 Comment

D4V1D Over a year ago

For sure, I didn't correct all the mistakes, just the one pointed out in the question :)

Michael J. Anderson · Accepted Answer · 2013-11-13 14:41:17Z

1

**echo "<name='TripID' id='TripID'>" . $check['TripID'] . " ";**

Looks like that should be:

echo "<input type='text' name='TripID' id='TripID' value='" . $check['TripID'] . "' />";

If you don't want it to be editable, display it then add a hidden field:

echo $check['TripID'];
echo "<input type='hidden' name='TripID' id='TripID' value='" . $check['TripID'] . "' />";

Basically, you're not putting your trip id into an actual form tag, so it's not getting posted over to your LastPage.php.

Edit: fixed the first input to wrap the tripID in the value attribute.

edited Nov 13, 2013 at 14:41

answered Nov 13, 2013 at 14:16

Michael J. Anderson

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4 Comments

Michael J. Anderson Over a year ago

Vinoth beat me to it. :)

D4V1D Over a year ago

I have always used value="myValue" to set up a default value to an input. Does <input type="text">myValue</input> as you suggested produce the same effects and is correct?

Daniel W. Over a year ago

@destroydaworld no it's invalid HTML, input is a void element dev.w3.org/html5/markup/input.text.html#input.text

Michael J. Anderson Over a year ago

Yeah, I had the input fixed in the second block, but missed it in the first block. correct syntax is <input value="" />. Going to update it now.

Himanshu Sharma · Accepted Answer · 2013-11-13 14:27:06Z

0

Replace following lines

**echo "<name='TripID' id='TripID'>" . $check['TripID'] . " ";**

with

echo "<input type="hidden" name="TripID" value="'.$check['TripID'].'"  />";

So it will not be visible to user on current page but when you will post the Form, it will be available in $_POST['TripID'] variable.

One more thing your tag is not properly closed.

Use MySQLi and prepared statements to prevent SQL injection.

PS: accept the answer if its work for you.

answered Nov 13, 2013 at 14:27

Himanshu Sharma

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Comments

Vinoth Babu · Accepted Answer · 2013-11-13 14:18:09Z

0

Your form tag is not closed properly. It is now as echo </form";

It should be echo"</form>";

Also you didnt added name in input tag.

It should be

echo"<input type='submit' value="$check['TripID']" name="TripID" class='Log'>";

There is no name tag <name> but you used how?

edited Nov 13, 2013 at 14:18

answered Nov 13, 2013 at 14:12

Vinoth Babu

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2 Comments

Daniel W. Over a year ago

echo" blankspace missing echo "; "$check['TripID']" dots missing " . $check['TripID'] . " You have errors in the answer omg don't advice people with wrong syntax

Gaston Velarde Over a year ago

Thank you vinoth for spotting the missing ">" in form. :)

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