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drop down menu showing old records from the database

im doing this project, they is a page called insert a review, in my insert review page , before you can write a review you select the title of the book, i use a drop down menu for this, so the user writes a review on an existing book in my books database,

but now the issue is, my drop down menu is showing old records, book's title that i have deleted. and when i display my books table, the old records don't show, which is correct.

i just want to know why is my drop down menu showing old deleted records?

<?php
echo "<select name='title' >";
$res=mysql_query("SELECT *  FROM mbooks");

while($entry=mysql_fetch_array($res))
{
    echo "<option";

    if($entry['title'] ==fred)
    {
        echo " selected=selected";
     } 

    echo ">"  .$entry['title']. "</option>";
}
echo "</select>";
?>

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Edited to put the code into the code snippet ~Ray
dropdownreviews.txt
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JCWEBHOST
Asked:
JCWEBHOST
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1 Solution
 
Chris StanyonCommented:
It absolutely impossible to tell from that.

Database records aren't cached, so you need to double check you're using the correct connection to the correct database, and using the correct table. Do you have the database on a development and a live server - make sure you select the right one.

Other than that, we're working blind here - nothing wrong with your code (apart from mySQL library being deprecated in PHP, but that won't affect the records!)
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Ray PaseurCommented:
We would need to see your CREATE TABLE statements and your test data model.  We might need to see the other processes that led to the current state of the test data model.  In other words, it's not a question with any one answer.  It's a research project.

Maybe step back from this a little bit and try to get a foundation in how PHP works?  This article can lead you to good learning resources.
http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/A_11769-And-by-the-way-I-am-new-to-PHP.html

And this article can help you avoid many of the bad things that happen to people who start copying code samples from the internet without any real understand of what they're doing.
http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/A_12293-AntiPHPatterns-and-AntiPHPractices.html

Since you will have to convert the data base extension from MySQL to something else, you might as well learn the right way now, and never go back to MySQL.
http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/PHP_Databases/A_11177-PHP-MySQL-Deprecated-as-of-PHP-5-5-0.html

Line 9 of the code snippet is almost certainly wrong, unless you have a defined constant somewhere else named "fred."  You might want to read this article, paying particular attention the part about time bombs.
http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/A_12241-Quotation-Marks-in-PHP.html
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JCWEBHOSTAuthor Commented:
what do you mean by  "deprecated in PHP" is they a new way to do a drop down menu?

my connect seems alright, because my other drop downs are working fine.
0
 
Chris StanyonCommented:
In PHP there are several libraries you can use to access your mySQL database - mysql, mysqli, and PDO - the mysql libraries are deprecated meaning that you will have to switch to using one of the other 2.

In your code you are using the mysql library (functions such as mysql_query() and mysql_fectch_array()). Read through Ray's article and it'll show you how to switch to using one of the other.

That said, it's not the cause of your problem, and switching to another library won't help you solve it (but it still needs to be done anyway)
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