function issue..

Question

geneh23 0 Junior Poster

12 Years Ago

Hey everyone! :)

I have a line of code that I am not sure about and I get an error message that says,
"Fatal error: Call to undefined function mysql_results() in C:\Program Files (x86)\EasyPHP-5.3.8.1\www\test5\func\blog.php on line 33"

when the code is from here in this script:

blog.php

<?php
function add_post($title, $contents, $category) {
	
}

function edit_post($id, $title, $contents, $category){
	
}

function add_category($name){
	$name = mysql_real_escape_string($name);
	
	mysql_query("INSERT INTO 'categories' SET 'name' = '$name'");
}

function delete($field, $id){
	
}

function get_posts($id = null, $cat_id = null){
	
}

function get_categories($id = null){
	
}

function category_exists($name){
	$name = mysql_real_escape_string($name);
	
	$query = mysql_query("SELECT COUNT(1) FROM `categories` WHERE `name` = `$name`");
	
	return ( mysql_results($query, 0) == 0 ) ? false : true;
}

the code that is in question is here:

return ( mysql_results($query, 0) == 0 ) ? false : true;

any help would be greatly appreciated! :) Thanks!

php

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Latest Post 12 Years Ago Latest Post by geneh23

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jogesh_p 6 Junior Poster in Training

12 Years Ago

the error is on the return line that is

return ( mysql_results($query, 0) == 0 ) ? false : true;

should be :

$return = ( mysql_result($query, 0) == 0 ) ? false : true;
return $return;

Edited 12 Years Ago by jogesh_p because: n/a

blocblue 238 Posting Pro in Training

12 Years Ago

Sorry, my mistake. What you had originally was very nearly correct.

function category_exists($name){
    $name = mysql_real_escape_string($name);
    $query = mysql_query("SELECT COUNT(*) FROM `categories` WHERE `name` = '$name'");
    return ! (mysql_result($query, 0) == 0);
}

The function should by mysql_result and checking that the count value is not equal to zero would determine whether the category exists.

R.

blocblue 238 Posting Pro in Training

12 Years Ago

@geneh23 - Please repost your code, indicating which line is line 33, because I have tested the function that I wrote and it works as expected for me, without a fatal error.

function category_exists($name){
    $name = mysql_real_escape_string($name);
    $query = mysql_query("SELECT COUNT(*) FROM `categories` WHERE `name` = '$name'");
    return ! (mysql_result($query, 0) == 0);
}

R.

blocblue 238 Posting Pro in Training

12 Years Ago

You've not copied my code correctly. Line 33 should read:

// Notice the double ==
return ! (mysql_result($query,0) == 0);

R

blocblue 238 Posting Pro in Training

12 Years Ago

You not copied the previous line correctly either. You have backticks around the name value. They should be single quotation marks instead. Replace your entire category_exists function with the following (copy and paste):

function category_exists($name){
    $name = mysql_real_escape_string($name);
    $query = mysql_query("SELECT COUNT(*) FROM `categories` WHERE `name` = '$name'");
    return ! (mysql_result($query, 0) == 0);
}

R.

geneh23 commented: this person know's his php! +3

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geneh23 0 Junior Poster · Answer 1 · 2011-11-19T14:36:36+00:00

@jogesh p: I did that and this error showed up:
"Fatal error: Call to undefined function mysql_results() in C:\Program Files (x86)\EasyPHP-5.3.8.1\www\test5\func\blog.php on line 33"

blocblue 238 Posting Pro in Training Featured Poster · Answer 2 · 2011-11-19T15:03:23+00:00

As @jogesh pointed out, there is no such function as mysql_results. Drop the s - mysql_result.

Also, mysql_result only returns false on error, otherwise it returns a MySQL result object, even if no rows are found. You would therefore need to use the mysql_num_rows function to determine whether no results were found.

function category_exists($name){
    $name = mysql_real_escape_string($name);
    $query = mysql_query("SELECT COUNT(*) FROM `categories` WHERE `name` = '$name'");
    $result = mysql_result($query);
    return (bool)mysql_num_rows($result);
}

R.

geneh23 0 Junior Poster · Answer 3 · 2011-11-19T15:16:13+00:00

@blocblue: when I changed the code to what you have, it comes up with these errors:

"Warning: mysql_result() expects at least 2 parameters, 1 given in C:\Program Files (x86)\EasyPHP-5.3.8.1\www\test5\func\blog.php on line 33"

"Warning: mysql_num_rows() expects parameter 1 to be resource, null given in C:\Program Files (x86)\EasyPHP-5.3.8.1\www\test5\func\blog.php on line 34"

:/

geneh23 0 Junior Poster · Answer 4 · 2011-11-20T03:05:18+00:00

@blocblue: now it says "Fatal error: Can't use function return value in write context in C:\Program Files (x86)\EasyPHP-5.3.8.1\www\test5\func\blog.php on line 33"

..I'm getting really tired of these errors.. :/

Smeagel13 5 Junior Poster in Training · Answer 5 · 2011-11-20T07:29:39+00:00

blocblue's showing off :P it would be simpler to use ...

return mysql_result($query, 0);

... providing the caller can handle FALSE if it's returned.

I'm hoping you've used if(category_exists('blahblah')) ...

geneh23 0 Junior Poster · Answer 6 · 2011-11-20T11:47:02+00:00

@Smeagel: Sorry for the lack of coding knowledge..do you mean like this?

if function category_exists($name){
	$name = mysql_real_escape_string($name);
	
	$query = mysql_query("SELECT COUNT(*) FROM `categories` WHERE `name` = `$name`");
	
	return mysql_result($query,0);
}

basically, I want this function to do as follows: "if 0 rows were returned with category name, return false. otherwise return true"

does that help?

geneh23 0 Junior Poster · Answer 7 · 2011-11-20T14:14:07+00:00

@blocblue: here is the entire code for blog.php

<?php
function add_post($title, $contents, $category) {

}

function edit_post($id, $title, $contents, $category){

}

function add_category($name){
    $name = mysql_real_escape_string($name);

    mysql_query("INSERT INTO 'categories' SET 'name' = '$name'");
}

function delete($field, $id){

}

function get_posts($id = null, $cat_id = null){

}

function get_categories($id = null){

}

function category_exists($name){
    $name = mysql_real_escape_string($name);

    $query = mysql_query("SELECT COUNT(*) FROM `categories` WHERE `name` = `$name`");

    return ! (mysql_result($query,0) = 0);
}

here is where there error is on line 33:

return ! (mysql_result($query,0) = 0);

geneh23 0 Junior Poster · Answer 8 · 2011-11-20T14:15:10+00:00

<?php
function add_post($title, $contents, $category) {
	
}

function edit_post($id, $title, $contents, $category){
	
}

function add_category($name){
	$name = mysql_real_escape_string($name);
	
	mysql_query("INSERT INTO 'categories' SET 'name' = '$name'");
}

function delete($field, $id){
	
}

function get_posts($id = null, $cat_id = null){
	
}

function get_categories($id = null){
	
}

function category_exists($name){
	$name = mysql_real_escape_string($name);
	
	$query = mysql_query("SELECT COUNT(*) FROM `categories` WHERE `name` = `$name`");
	
	return ! (mysql_result($query,0) = 0);
}

all of code^^^

geneh23 0 Junior Poster · Answer 9 · 2011-11-20T14:20:05+00:00

@blocblue here is all of the code in blog.php

<?php
function add_post($title, $contents, $category) {
	
}

function edit_post($id, $title, $contents, $category){
	
}

function add_category($name){
	$name = mysql_real_escape_string($name);
	
	mysql_query("INSERT INTO 'categories' SET 'name' = '$name'");
}

function delete($field, $id){
	
}

function get_posts($id = null, $cat_id = null){
	
}

function get_categories($id = null){
	
}

function category_exists($name){
	$name = mysql_real_escape_string($name);
	
	$query = mysql_query("SELECT COUNT(*) FROM `categories` WHERE `name` = `$name`");
	
	return ! (mysql_result($query,0) = 0);
}

and here is where the issue is on line 33

return ! (mysql_result($query,0) = 0);

geneh23 0 Junior Poster · Answer 10 · 2011-11-20T14:21:03+00:00

geneh23 0 Junior Poster

12 Years Ago

sorry for the repost..but there it is :)

blocblue 238 Posting Pro in Training Featured Poster · Answer 11 · 2011-11-20T14:22:52+00:00

You've not copied my code correctly. Line 33 should read:

// Notice the double ==
return ! (mysql_result($query,0) == 0);

R

geneh23 0 Junior Poster · Answer 12 · 2011-11-20T14:32:33+00:00

@blocblue: I'm sorry, I misread what you put..I added the extra "=" however, now an error shows this:
"Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\Program Files (x86)\EasyPHP-5.3.8.1\www\test5\func\blog.php on line 33"

geneh23 0 Junior Poster · Answer 13 · 2011-11-20T15:36:14+00:00

@blocblue: You are right! and there were a couple of other small issues that I noticed. For the mysql queries..in blog.php ...there were " ' " instead of " ` " and I actually needed to add a one to SELECT COUNT () because It's checking to see if there is a record of something in the database and it can't check for something that isn't in there..so therefore it would return false..but you're code helped a lot. As always! Thanks for your help! I'll do the up arrow thing on every post..the same goes for everyone else who attempted to help me :) SOLVED!!

function issue..

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