PHP to mySQL Connection

mkebkr04 · November 14, 2008

I am trying to link my php site to a mySQL database. So far it seems as though everything connects right because when I enter information into the form it says successful. I then check the database and it did indeed create a new row but only of blanks, 1s, 0s, and NULLs. So I am kind of at a stand point right now.

<?

if (isset($_POST['submit']))

{

$business = $_POST['Business_name'];

$title = $_POST['Job_title'];

$salary = $_POST['Salary'];

$cname = $_POST['Business_contact_name'];

$cemail = $_POST['Business_contact_email'];

$cphone1 = $_POST['Business_contact_phone1'];

$cphone2 = $_POST['Business_contact_phone2'];

$business_url = $_POST['Business_url'];

$intern_url = $_POST['Intern_url'];

$fname = $_POST['Faculty_contact_name'];

$femail = $_POST['Faculty_contact_email'];

$fphone = $_POST['Faculty_contact_phone'];

$username = $_POST['Username'];

$sdate = $_POST['Internship_start_date'];

$edate = $_POST['Internship_end_date'];

$exdate = $_POST['Expiration_date'];

$sdescription = $_POST['Short_description'];

$description = $_POST['Description'];

$con = mysql_connect("host", "username", "pass");

if (!$con)

{

die('Could not connect: ' . mysql_error());

}

mysql_select_db("bakerjm", $con);

$sql = "INSERT INTO Interns(

Business_name, Job_title, Salary, Business_contact_name, Business_contact_email, Business_contact_phone1,

Business_contact_phone2, Business_url, Internship_url, Faculty_contact_name, Faculty_contact_email,

Faculty_contact_phone, Username, Internship_start_date, Internship_end_date, Expiration_date,

Short_description, Description)

VALUES('$business','$title','$salary','$cname','$cemail','$cphone1','$cphone2','$business_url','$intern_url','$fname',

'$femail','$fphone','$username','$sdate','$edate','$exdate','$sdescription','$description')";

if (!mysql_query($sql, $con))

{

die('Error: ' . mysql_error());

}

echo "Submitted Successfully";

mysql_close($con);

}

?>

That is what I am using as of right now. The inputs are mostly text except for a few drop down menus that include dates.

Thanks in advance for any help.

revraz · November 14, 2008

echo $sql and see what it contains.

mkebkr04 · November 14, 2008

This is what it displays after using echo $sql but this is my first php project and I am unsure if this is supposed to be like this or not. I am guessing that VALUES is supposed to actually have variables between the quotation marks?

Submitted SuccessfullyINSERT INTO Interns( Business_name, Job_title, Salary, Business_contact_name, Business_contact_email, Business_contact_phone1, Business_contact_phone2, Business_url, Internship_url, Faculty_contact_name, Faculty_contact_email, Faculty_contact_phone, Username, Internship_start_date, Internship_end_date, Expiration_date, Short_description, Description) VALUES('','','','','','','','','','', '','','','','','','','')

Jibberish · November 14, 2008

in the <form> tag from the page that sends the data what method are you using? POST or GET?

PFMaBiSmAd · November 14, 2008

Either your form is invalid or the names of the form fields do not match what you are using in the php code.

At least the submit field in your form is working. Have you checked that the names of your form fields match what you are using in the php code?

When learning php, developing php code, or debugging php code, set error_reporting to E_ALL and set display_errors to ON. These settings should be done in your php.ini (stop and start your web server to get any changes made to php.ini to take effect) so that you will also get fatal parse errors to be displayed.

mkebkr04 · November 14, 2008

The form method is post and the fields all match. There is one thing that might be causing it possibly. I have drop down menus for three dates so nine menus total. The format is January 1 2008. So for example the name of one of these dates is Expiration_date, what would I use in this case....DATETIME, just DATE or would I use something else on the mySQL side to take in the date from the PHP side. If this could be the problem would it cause the whole form to fail like it is?

PFMaBiSmAd · November 14, 2008

Except for your submit field, none of your form variables have any values. Either your form is invalid or the form field names don't match. You would need to post your form to get any specific help with why it is not submitting data to your form processing code.

mkebkr04 · November 14, 2008

The more I look at it the more I think the date part is messed up.

<form action="<?= $_SERVER['PHP_SELF'] ?>" method="post">

<br>

<font size="+1">Add New Intern/Job Posting</font>

Please complete the following fields

<tr>

<td align="left" valign="bottom"><p>Name of Business: </p></td>

</tr>

<tr>

<td align="left" valign="bottom"><p>Job Title: </p></td>

</tr>

<tr>

<td align="left" valign="bottom"><p>Business Contact Name: </p></td>

</tr>

<tr>

<td align="left" valign="bottom"><p>Business Contact Email: </p></td>

</tr>

<tr>

<td align="left" valign="bottom"><p>Business URL: </p></td>

</tr>

<tr>

<td align="left" valign="bottom"><p>Internship URL: </p></td>

</tr>

<tr>

<td align="left" valign="bottom"><p>Faculty Contact Name: </p></td>

</tr>

<tr>

<td align="left" valign="bottom"><p>Faculty Contact Email: </p></td>

</tr>

<tr>

<td align="left" valign="bottom"><p>Submitted by (Username): </p></td>

</tr>

<tr>

<td align="left" valign="bottom"><p>Business Contact Phone #1: </p></td>

</tr>

<tr>

<td align="left" valign="bottom"><p>Business Contact Phone #2: </p></td>

</tr>

<tr>

<td align="left" valign="bottom"><p>Faculty Contact Phone: </p></td>

</tr>

<tr>

<td align="left" valign="bottom"><p>Salary: </p></td>

</tr>

<tr>

Internship Start Date: </p></td>

<?

for($i=1;$i<=12;$i++)

{

echo "<option value='$i'>$months[$i]</option>\n";

}

?>

</select>

<?

for($i=1;$i<=31;$i++)

{

echo "<option value='$i'>$i</option>\n";

}

?>

</select>

<?

$current_year = date("Y");

for($i=$current_year-1;$i<=$current_year+2;$i++)

{

if( $i == $current_year )

{

echo "<option SELECTED value='$i'>$i\n";

} else

{

echo "<option value='$i'>$i\n";

}

?>

</select>

</p></td>

</tr>

<tr>

Internship End Date: </p></td>

<?

for($i=1;$i<=12;$i++)

{

echo "<option value='$i'>$months[$i]</option>\n";

}

?>

</select>

<?

for($i=1;$i<=31;$i++)

{

echo "<option value='$i'>$i</option>\n";

}

?>

</select>

<?

$current_year = date("Y");

for($i=$current_year-1;$i<=$current_year+2;$i++)

{

if( $i == $current_year )

{

echo "<option SELECTED value='$i'>$i\n";

} else

{

echo "<option value='$i'>$i\n";

}

?>

</select>

</p></td>

</tr>

<tr>

Expiration Date: </p></td>

<?

for($i=1;$i<=12;$i++)

{

echo "<option value='$i'>$months[$i]</option>\n";

}

?>

</select>

<?

for($i=1;$i<=31;$i++)

{

echo "<option value='$i'>$i</option>\n";

}

?>

</select>

<?

$current_year = date("Y");

for($i=$current_year-1;$i<=$current_year+2;$i++)

{

if( $i == $current_year )

{

echo "<option SELECTED value='$i'>$i\n";

}

else

{

echo "<option value='$i'>$i\n";

}

?>

</select>

</p></td>

</tr>

</table>

<p>

</p>

<table>

<td align="center"><p>Short Description:

</p></td>

</tr>

<td align="center"><p>Description:

</p></td>

</tr>

</table>

</form>

PFMaBiSmAd · November 14, 2008

NONE of your form's data fields match the $_POST['....'] variable names you are using.

runnerjp · November 14, 2008

and use code tags!!

<php echo ' the code tags will make it easyer to read?!?!?! ?>

Jibberish · November 14, 2008

the name of the input box has to be the same as the $_POST[].

you have things like

<input type="text" name="business" size="50" maxlength="50" />

and the post as

<?php
$business = $_POST['Business_name'];

?>

and it should be

<?php
$business = $_POST['business'];

?>

they have to match in order to get them

nitation · November 14, 2008

follow Jibberish last post and i believe your problem will be history!.

My 2 cent

mkebkr04 · November 14, 2008

Thanks alot it works perfectly now, except for one thing. Say I have start_month, start_day, start_year and I want all three to be put into one column in the database. I tried $exdate = $_POST['exp_month','exp_day','exp_year']; and its complains about the commas?? Any ideas??

mkebkr04 · November 14, 2008

Say I have start_month, start_day, start_year and I want all three to be put into one column in the database. I tried $exdate = $_POST['exp_month','exp_day','exp_year']; and its complains about the commas?? Any ideas??

wildteen88 · November 14, 2008

Look into using the concatenation operator (.), example

$var1 = 'Hello';
$var2 = 'World';

echo $var1 . ', ' . $var2;

So with your example you'd use

$exdate = $_POST['exp_month'] . $_POST['exp_day'] . $_POST['exp_year'];

mkebkr04 · November 14, 2008

That definitly makes sense but it doesn't seem to want to work. Do I need some kind of format function or something for the date to go into my SQL. After I hit submit it shows dates as follows: 0000-00-00, so does php accomadate this order or would I need to put it in that order somehow?

wildteen88 · November 14, 2008

No this is due to how you have setup your table.

You have set your date columns datatype as date which expects the following date format yyyy-mm-dd. However your date format is mmddyyyy. You should change this

$exdate = $_POST['exp_month'] . $_POST['exp_day'] . $_POST['exp_year'];

to

$exdate = sprintf('%d-%02d-%02d', $_POST['exp_year'], $_POST['exp_month'], $_POST['exp_day']);

mkebkr04 · November 19, 2008

Thanks, you have all been a lot of help now I am moving from the create.php to the edit.php. Les just say hopefully I learn a lot from this. Probably more questions to come. Thanks again

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