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debugging assistance...


nueamin

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Thank You in Advance. I am rather frustrated at this juncture. Ive spent hours trying to work this one out. I also apologize completely for any of my poor coding standards. I would say i am self taught but I am actually learning off of forums like these. I have checked and rechecked the user name and password and it has all permissions. The echo $query produces a line which runs in the MySQL query window flawlessly. Is there something I am doing wrong or something I should look into. Thank You everyone for your time and help. Oh yea. What is wrong? The data that was input via form is not being input into the database, although the query echo shows the data in the string.

 

$db_username="nottelling";

$db_password="reallynottelling";

$database="realmof2_profiles";

 

$cfirst_name=$_POST['ud_first_name'];

$cmiddle_init=$_POST['ud_middle_init'];

$clast_name=$_POST['ud_last_name'];

$ce_mail=$_POST['ud_e_mail'];

$cphone=$_POST['ud_phone'];

$cstreet_ad=$_POST['ud_street_ad'];

$ccity=$_POST['ud_city'];

$cstate=$_POST['ud_state'];

$czipcode=$_POST['ud_zipcode'];

 

mysql_connect(localhost,$db_username,$db_password);

$query = sprintf("UPDATE contactinfo SET first_name='$cfirst_name', middle_init='$cmiddle_init', last_name='$clast_name', e_mail='$ce_mail',phone='$cphone', street_ad='$cstreet_ad', city='$ccity', state='$cstate', zipcode='$czipcode' WHERE user_id='$userid'");

echo $query;

mysql_query($query);

mysql_close();

echo '  </div>';

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Try adding the mysql_error() function like I've written below. It will print a database error in your web browser when you try to run the code.

$db_username="nottelling";
$db_password="reallynottelling";
$database="realmof2_profiles";

$cfirst_name=$_POST['ud_first_name'];
$cmiddle_init=$_POST['ud_middle_init'];
$clast_name=$_POST['ud_last_name'];
$ce_mail=$_POST['ud_e_mail'];
$cphone=$_POST['ud_phone'];
$cstreet_ad=$_POST['ud_street_ad'];
$ccity=$_POST['ud_city'];
$cstate=$_POST['ud_state'];
$czipcode=$_POST['ud_zipcode'];

$link = mysql_connect(localhost,$db_username,$db_password);
$query = sprintf("UPDATE contactinfo SET first_name='$cfirst_name', middle_init='$cmiddle_init', last_name='$clast_name', e_mail='$ce_mail',phone='$cphone', street_ad='$cstreet_ad', city='$ccity', state='$cstate', zipcode='$czipcode' WHERE user_id='$userid'");
echo $query;
mysql_query($query) or die mysql_error($link);
mysql_close();
echo '   </div>';

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What are you actually trying to do here? And what is it doing instead?

 

(Are you trying to insert a new field or update an existing field?)

 

Try this after running the query, but before you close the connection:

echo $mysql_error();

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