Query MySQL to show users post in PHP - Not sure how to formulate query

Question

This is my existing code from a tutorial:

<fieldset>
    <legend><?php echo WORDING_EDIT_USER_DATA; ?></legend>
    <p><?php echo WORDING_YOU_ARE_LOGGED_IN_AS . '<b>' . $_SESSION['user_name']; ?></b></p><hr/>

    <p>$_SESSION['user_id'] = <?php echo $_SESSION['user_id']; ?></p>
    <p>$_SESSION['user_name'] = <?php echo $_SESSION['user_name']; ?></p>
    <p>$_SESSION['user_email'] = <?php echo $_SESSION['user_email']; ?></p>
    <p>$_SESSION['user_access_level'] = <?php echo $_SESSION['user_access_level']; ?></p>
    <p>$_SESSION['user_logged_in'] = <?php echo $_SESSION['user_logged_in']; ?></p><hr/>

    <p><?php echo WORDING_PROFILE_PICTURE . '<br/><img src="' . $login->getGravatarImageUrl() ; ?>" /></p>
</fieldset><br/>
<a href="?logout"><?php echo WORDING_LOGOUT; ?></a>&nbsp;&nbsp;|&nbsp;&nbsp;<a href="?edit"><?php echo WORDING_EDIT_USER_DATA; ?></a>
<?php echo (ALLOW_ADMIN_TO_REGISTER_NEW_USER && $_SESSION['user_access_level'] == 255 ? '<br/><a href="?register">'. WORDING_REGISTER_NEW_ACCOUNT .'</a>' : ''); ?>

This is the query I want to do on this schema:

SELECT item_title, item_location, item_datetime
FROM item
WHERE user_id = 1;

I'm fairly new to this but this is what I was trying, in order to show the post items which belong to that user who is logged in:

<?php

$result = mysql_query('SELECT item_title, item_location, item_datetime
FROM item WHERE user_id = $_SESSION['user_id'] ;');
if (!$result) {
    die('Invalid query: ' . mysql_error());
}

?>

I haven't been able to make much progress in making this query work or this code run. Can anyone tell me where I am going wrong please?

if you are getting mysql error then try to print your query and check if user id is showing or not — dev.php
– dev.php, Commented Feb 2, 2016 at 7:43
check your quotation marks, your query should be like "SELECT item_title, item_location, item_datetime FROM item WHERE user_id = $_SESSION['user_id'] ;" — Luthando Ntsekwa
– Luthando Ntsekwa, Commented Feb 2, 2016 at 7:43

Luthando Ntsekwa · Accepted Answer · 2016-02-02 07:44:16Z

2

Replace the main boundaries with double qoutes

mysql_query('SELECT item_title, item_location, item_datetime
FROM item WHERE user_id = $_SESSION['user_id'] ;');

to

mysql_query("SELECT item_title, item_location, item_datetime
FROM item WHERE user_id = $_SESSION['user_id'] ;");

edited Feb 2, 2016 at 7:44

Luthando Ntsekwa

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answered Feb 2, 2016 at 7:43

erfjax

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2 Comments

Jimmy Over a year ago

How come the other answer has a full stop in front of the $_SESSION and yours doesn't? Does it make a difference?

Luthando Ntsekwa Over a year ago

@Jimmy ask google....Double quote strings displays a host of escaped characters (including some regexes), and variables in the strings are evaluated. Use double quotes in PHP to avoid having to use the period to separate code.

Berserk · Accepted Answer · 2016-02-02 07:43:03Z

2

Replace your query by this one and try :

$result = mysql_query('SELECT item_title, item_location, item_datetime
FROM item WHERE user_id ='.$_SESSION['user_id']);

answered Feb 2, 2016 at 7:43

Berserk

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Comments

Luthando Ntsekwa · Accepted Answer · 2016-02-02 07:52:16Z

1

try this:

<?php

$result = mysql_query("SELECT item_title, item_location, item_datetime
FROM item WHERE user_id =  " .$_SESSION['user_id']. " ;");
if (!$result) {
    die('Invalid query: ' . mysql_error());
}

?>

Anyway, you should escape the variable $_SESSION['user_id'], to prevent SQL injection.

Here is some of the official documentation:

http://php.net/manual/es/security.database.sql-injection.php

edited Feb 2, 2016 at 7:52

Luthando Ntsekwa

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answered Feb 2, 2016 at 7:46

Peter Black Moore

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