MySQL not fetching correct data? (PHP)

Question

This is pretty straight forward. EDIT: Updated question and added fourth echo.

Here is PHP code:

<?php

$ratings="3";
$item="Inception";

$query="SELECT * FROM items WHERE item = '". $item ."' LIMIT 1";

echo $query;

echo "<br />";

$result=mysql_query($query);

echo $result;

echo "<br />";

while ($row = mysql_fetch_array($result)) { 
    $item_id = $row['item_id'];
    echo $item_id;
    echo "<br />";
 }  

  $query_two = "INSERT INTO ratings (rating, item_id), VALUES (' {$ratings} ', ' {$item_id} ')";

  echo $query_two;
  $sql = mysql_query($query_two);
  mysql_close();
?>

Here is web output with all the echo's:

SELECT * FROM items WHERE item = 'Inception' LIMIT 1
Resource id #7


INSERT INTO ratings (rating, item_id), VALUES (' 3 ', ' ')

How come my $item_id is blank? (third row underneath Resource id)

That resource is from echo $result;, what did you think would happen there? — Wrikken
– Wrikken, Commented Jun 20, 2012 at 16:18
I was asking question from the wrong angle. And just answered my own question. Why is $items_id blank? because it should be $item_id -- a typo. But everyone did answer my original question. Thanks! — TheLettuceMaster
– TheLettuceMaster, Commented Jun 20, 2012 at 16:27

Zbigniew · Accepted Answer · 2012-06-20 16:23:41Z

This part of code produces it:

$result=mysql_query($query);

echo $result;

It shows Resource... because it is of resource type, it's just a sort of special handler for query, it's not like normal type (string or int for example), so it has nothing readable to print.

If you want to print data from query then you must firstly fetch it.

Also note that those mysql_* functions are deprecated, it is discouraged to use them. Note from php manual:

Use of this extension is discouraged. Instead, the MySQLi or PDO_MySQL extension should be used. See also MySQL: choosing an API guide and related FAQ for more information. Alternatives to this function include:

mysqli_query()

PDO::query()

poncha · Accepted Answer · 2012-06-20 16:17:51Z

4

This does not have anything to do with IDs from the database.

This (Result#7) says that this result resource is seventh resource to be created by your php script execution.

answered Jun 20, 2012 at 16:17

poncha

7,8763 gold badges36 silver badges39 bronze badges

Comments

Fahim Parkar · Accepted Answer · 2012-06-20 16:20:44Z

3

Also

$query_two = "INSERT INTO ratings (rating, item_id), VALUES (' {$ratings} ', ' {$item_id} ')";

should be

$query_two = "INSERT INTO ratings (rating, item_id) VALUES (' {$ratings} ', ' {$item_id} ')";

You have comma before VALUES.

Also, it seems that $item_id is blank. Please check DB whether you have data for item = 'Inception'.

Regarding, Result#7 please follow others answers.

answered Jun 20, 2012 at 16:20

Fahim Parkar

1

1 Comment

bfavaretto Over a year ago

Another possible reason for the blank item_id: there may be no rows matching the query condition.

Sandeep Bansal · Accepted Answer · 2012-06-20 16:19:01Z

2

The Resource ID is coming from the actual process/object that the MySQL Query is.

to return the result you need:

$row = mysql_fetch_array( $query );

echo $row['item']

answered Jun 20, 2012 at 16:19

Sandeep Bansal

6,44018 gold badges88 silver badges128 bronze badges

Comments

Andrew Kozak · Accepted Answer · 2012-06-20 16:25:57Z

2

You need to do something with the result resource. Try this:

$result=mysql_query($query);

//echo $result;

$result_array = mysql_fetch_assoc( $result );

print_r( $result_array );

EDIT: I see you updated your question.

You should run your item='Inception' query directly in MySQL to confirm that results are what you expect.

edited Jun 20, 2012 at 16:25

answered Jun 20, 2012 at 16:19

Andrew Kozak

1,6602 gold badges22 silver badges35 bronze badges

Comments

David · Accepted Answer · 2012-06-20 16:22:49Z

1

You cannot echo the result that simple. You need to fetch the result to for example an array:

while ($row = mysql_fetch_array($query)) {
  echo $row['a_column'] . "<br />";
}

or an object:

while ($variable = mysql_fetch_object($query) {
 $value = $variable->a_column;
}
echo $value;

There are more ways but this is just two examples

answered Jun 20, 2012 at 16:22

David

1,1918 gold badges26 silver badges49 bronze badges

Collectives™ on Stack Overflow

MySQL not fetching correct data? (PHP)

6 Answers 6

Comments

Comments

1 Comment

Comments

Comments

Comments

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

Comments

Comments

1 Comment

Comments

Comments

Comments

Related