Programming Forums > Web Development and Design > PHP

nickk · Nov 7th, 2004, 12:00 PM

Hi,

I have 2 questions:

Firstly, I have the following function, but it doesnt work Sad

function insert($school_id, $email, $text, $pic) {
 * * $sql = "select class, first_name, second_name from yearbook where school_id = '$school_id'";
 * * $result = mysql_query($sql);
 * * $class = mysql_result($result, 1, "class");
 * * $first_name = mysql_result($result, 1, "first_name");
 * * $last_name = mysql_result($result, 1, "last_name");
 * * $pic = "../" . $class . "/" . $first_name . "-" . $last_name . ".jpg";

 * * $sql = "update yearbook set email = '$email', text = '$text', pic = '$pic' "
    * . "where school_id = '$school_id'";
 * * if(mysql_query($sql)) {
 * * * * *echo "<b>Text and picture sucessfully uploaded!</b>\n";
 * * * * *return 1;
 * * } else {
 * * * * *echo "<b>Error submitting text and picture to the database, go back and try again</b>\n";
 * * * * *return 0;
 * * }
} //end function insert

It gives me the following errors:
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /usr/home/nickk/public_html/yearbook/provide/upload.php on line 228

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /usr/home/nickk/public_html/yearbook/provide/upload.php on line 229

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /usr/home/nickk/public_html/yearbook/provide/upload.php on line 230

and it doesnt add the proper values for "pic" to the database Sad

Secondly, I'm having problems uploading files (pictures to be exact) with PHP. Does anyone know how this is done?

Thanks, Nick

(added a little code formatting, courtesy of your friendly neighborhood Spider Man .. [ AKA: SykkN ])

BlazingWolf · Nov 7th, 2004, 1:08 PM

I can't help you with the picture uploading but I can tell say you are trying run an query without ever connecting to the databse

.

Before you run the query you need to have

(Toggle Plain Text)

$dbconnect = mysql_connect('localhost', 'root', 'mypassword');
mysql_select_db('mydb');

And when your run your query you need to include the vaible that has your connection info

(Toggle Plain Text)

$result = mysql_query($command,$dbconnect);

And as a good habit after you run your query you want to close the connection to the database.

(Toggle Plain Text)

mysql_close($dbconnect);

I can't help you with mysql_result because I've never used it before and I can't figure what your trying to do with it.

Ade · Nov 7th, 2004, 1:14 PM

I think you need to mysql_fetch_array($result)

tempest · Nov 7th, 2004, 2:15 PM

No, he doesnt need mysql_fetch_array or mysq_fetch_assoc but he can make it easier on himself by using it...

kurifu · Nov 8th, 2004, 2:04 AM

mysql_result is perfectly acceptable. Whenever you get an error such as that it means that your query was rejected by the server. You need to debug and figure out why, I would suggest using print_r( $query_handle ) or mysql_error( ) for more information.

In this case it looks to simply be that you forgot to connect to the database server. mysql_connect() is what you want for that.

Nov 7th, 2004, 12:00 PM	#1
nickk Newbie Join Date: Oct 2004 Posts: 4 Rep Power: 0	Hi, I have 2 questions: Firstly, I have the following function, but it doesnt work Sad (Toggle Plain Text) function insert($school_id, $email, $text, $pic) { * * $sql = "select class, first_name, second_name from yearbook where school_id = '$school_id'"; * * $result = mysql_query($sql); * * $class = mysql_result($result, 1, "class"); * * $first_name = mysql_result($result, 1, "first_name"); * * $last_name = mysql_result($result, 1, "last_name"); * * $pic = "../" . $class . "/" . $first_name . "-" . $last_name . ".jpg"; * * $sql = "update yearbook set email = '$email', text = '$text', pic = '$pic' " * . "where school_id = '$school_id'"; * * if(mysql_query($sql)) { * * * * echo "<b>Text and picture sucessfully uploaded!</b>\n"; * * * return 1; * } else { * * * * echo "<b>Error submitting text and picture to the database, go back and try again</b>\n"; * * * return 0; * } } //end function insert function insert($school_id, $email, $text, $pic) { * * $sql = "select class, first_name, second_name from yearbook where school_id = '$school_id'"; * * $result = mysql_query($sql); * * $class = mysql_result($result, 1, "class"); * * $first_name = mysql_result($result, 1, "first_name"); * * $last_name = mysql_result($result, 1, "last_name"); * * $pic = "../" . $class . "/" . $first_name . "-" . $last_name . ".jpg"; * * $sql = "update yearbook set email = '$email', text = '$text', pic = '$pic' " * . "where school_id = '$school_id'"; * * if(mysql_query($sql)) { * * * * echo "<b>Text and picture sucessfully uploaded!</b>\n"; * * * return 1; * } else { * * * * echo "<b>Error submitting text and picture to the database, go back and try again</b>\n"; * * * return 0; * } } //end function insert It gives me the following errors: Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /usr/home/nickk/public_html/yearbook/provide/upload.php on line 228 Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /usr/home/nickk/public_html/yearbook/provide/upload.php on line 229 Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /usr/home/nickk/public_html/yearbook/provide/upload.php on line 230 and it doesnt add the proper values for "pic" to the database Sad Secondly, I'm having problems uploading files (pictures to be exact) with PHP. Does anyone know how this is done? Thanks, Nick (added a little code formatting, courtesy of your friendly neighborhood Spider Man .. [ AKA: SykkN ])

Nov 7th, 2004, 1:08 PM	#2
BlazingWolf Hobbyist Programmer Join Date: Sep 2004 Posts: 207 Rep Power: 5	I can't help you with the picture uploading but I can tell say you are trying run an query without ever connecting to the databse . Before you run the query you need to have (Toggle Plain Text) $dbconnect = mysql_connect('localhost', 'root', 'mypassword'); mysql_select_db('mydb'); $dbconnect = mysql_connect('localhost', 'root', 'mypassword'); mysql_select_db('mydb'); And when your run your query you need to include the vaible that has your connection info (Toggle Plain Text) $result = mysql_query($command,$dbconnect); $result = mysql_query($command,$dbconnect); And as a good habit after you run your query you want to close the connection to the database. (Toggle Plain Text) mysql_close($dbconnect); mysql_close($dbconnect); I can't help you with mysql_result because I've never used it before and I can't figure what your trying to do with it. __________________ _______________________________ BlazingWolf

Nov 7th, 2004, 1:14 PM	#3
Ade Hobbyist Programmer Join Date: Oct 2004 Location: England, UK Posts: 139 Rep Power: 0	I think you need to mysql_fetch_array($result) __________________ Don't wound what you can't kill

Nov 7th, 2004, 2:15 PM	#4
tempest Programming Guru Join Date: Oct 2004 Posts: 1,041 Rep Power: 6	No, he doesnt need mysql_fetch_array or mysq_fetch_assoc but he can make it easier on himself by using it... __________________

Nov 8th, 2004, 2:04 AM	#5
kurifu Expert Programmer Join Date: Jul 2004 Location: Halifax, Nova Scotia (Canada) Posts: 784 Rep Power: 5	mysql_result is perfectly acceptable. Whenever you get an error such as that it means that your query was rejected by the server. You need to debug and figure out why, I would suggest using print_r( $query_handle ) or mysql_error( ) for more information. In this case it looks to simply be that you forgot to connect to the database server. mysql_connect() is what you want for that. __________________ Clifford Matthew Roche <geek@cliffordroche.com> Web Hosting: http://www.crd-hosting.com Consulting: http://www.crdev-consulting.com

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Programming Forums > Web Development and Design > PHP

2 Questions