What am I missing? Checkboxes

sideslip

I'm still learning my way round PHP and mySQL (apparently quite slowly :rolleyes: ) so I'd really appreciate any help or advice you might be able to give me as to why I can't get the following to work.

I have a table called "jobs" with 3 fields currently - "id", "jobtitle" and "jobdescrip"

I have a form to upload the info to the table, a form to edit it and a way of deleting it which all works nicely.

Problem is that I want to introduce a checkbox in the form so that I can "do stuff" with the results - like only show the results from rows where the checkbox was ticked. BUT I can't seem to pass the checkbox variable into the table at all.

Here's the uploading form:

<form action="addjob.php" method="post">
<input type="hidden" name="id" value="NULL" />
<input type="checkbox" id="show" name="show" value="1" />Show Job on Recruitment Page?
Job Title
<input type="text" name="jobtitle" title="input job title" id="jobtitle" size="50" />
Enter full Job Description
<textarea name="jobdescrip" id="jobdescrip" rows="15" cols="50"></textarea>
<input type="submit" value="Submit Job Details" />
</fieldset>
</form>

Here's the addjob.php page:

.............

/making variables
$jobtitle = $_POST['jobtitle'];
$jobdescrip = $_POST['jobdescrip'];
$show = $_POST['show'];

//oh the query!
$result=MYSQL_QUERY("INSERT INTO jobs (id,jobtitle,jobdescrip,show)".
"VALUES ('NULL', '$jobtitle', '$jobdescrip', '$show')");

.................

Any advice would be gratefully accepted

Edit: apologies if this should be in the newbies forum - please feel free to move if appropriate.

gardnc

checkbox posted is a true or false. You need to test it for true and save an appropriate value in the row value. i.e.

if ($_POST['show']) {
$show = "Y";
}

mourisj

I don't really agree with gardnc. With checkboxes you can submit any value, with several checkboxes with the same name even several values!

I'd suggest to check what your query really looks like.

$qry = "INSERT INTO jobs (id,jobtitle,jobdescrip,show)".
"VALUES ('NULL', '$jobtitle', '$jobdescrip', '$show')");

echo $qry;

and see if there is any complaint from mysql.

$result = mysql_query($qry)
or die(mysql_error());

Also try to submit the query contained in $qry directly to mysql (on comman-line tool or phpmyadmin or similar.

Hope this helps

JJ Mouris

gardnc

Well, I am talking out of school here, but I've never been able to save a value from a checkbox and the code I presented works for me.

sideslip

Thanks to both of you - I'll have a go at this in the morning when my brain's co-operating!

mourisj

I did not have any trouble, even with multiple values. You needthen to define the name as an array, like in this code which works fine for me:

echo "<p><input type=\"checkbox\" name=\"add_opt[]\" value=\"1\">Images définies</input></p>\n";
echo "<p><input type=\"checkbox\" name=\"add_opt[]\" value=\"2\">Images non définies</input></p>\n";
echo "<p><input type=\"checkbox\" name=\"add_opt[]\" value=\"3\">Pas de modification des anciennes images</input></p>\n";

To work with the data submitted, I do sth like:

$opt = $_POST[add_opt];
if (in_array(3, $opt) )
{
...
}

HTH

JJ Mouris

sideslip

Hi again

I've pretty much got this working now - the problem with the code that I posted here turned out to be that "show" is a protected word so I changed that field name and it works now.

Thanks again!