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I have code

$email = "[email protected]";    
$row = mysql_query("SELECT EXISTS(SELECT email FROM accounts WHERE email='".$email."');");
echo $row[0];

However, nothing is echoed.

This is strange because something is being returned because, later in the code I have a function that is CALLED:

if ( $row[0] == 0 ) { echo "Email address is available<br>";};

However: this is strange because when i put the SAME CODE into mySQL database command prompt:

enter image description here

It clearly returns 1 or TRUE.

The mysql_query is returning 0 when the same exact command in mysql command prompt returns 1. Further: I am unable to echo the result for debugging purposes.

EDIT: Please not, the regular mySQL command is returning this and ONLY this: enter image description here

EDIT: Here is there entire database: enter image description here

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2
  • have you tried doing print_r($row) to see what all is being returned? Commented May 28, 2014 at 20:51
  • Yeah that just gives me a Resource id #5 Commented May 28, 2014 at 20:52

3 Answers 3

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3

MySQL query gives you a ressource. After that you have to fetch the data with mysql_fetch_assoc or mysql_fetch_row or something else for example. But its better to use prepared statements with mysqli or PDO to get more security.

$email = "[email protected]";    
$res = mysql_query("SELECT EXISTS(SELECT email FROM accounts WHERE email='".myql_real_escape_string($email)."')");
$row = mysql_fetch_assoc($res);
echo $row['email'];

Answer to your question: 

$email = "[email protected]";    
$res = mysql_query("SELECT email FROM accounts WHERE email='".mysql_real_escape_string($email)."')");
$numRows = mysql_num_rows($res);

if($rowRows > 0) {
    echo "Record Available";
}
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7 Comments

Since i'm doing SELECT EXISTS its only going to return true or false right?
Hmm if you want to check if the data exists its easier to make a count SELECT COUNT(id) as countout FROM ... WHERE ... if row['countout'] is bigger then 0 its a dataset available.
I tried $email = "[email protected]"; $res = mysql_query("SELECT EXISTS(SELECT email FROM accounts WHERE email='".$email."')"); $row = mysql_fetch_assoc($res); echo $row['email']; However i got errors: Notice: Undefined index: email in /Applications/XAMPP/xamppfiles/htdocs/TLB/create_account.php
Your second code snippet is giving me error: Notice: Undefined index: countout in
Try that with mysql_num_rows then you get the number of all counted rows.
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0

You need to actually retrieve the result set from the query. mysql_query() just returns a resource handle for a successful select. You then need to fetch the results using mysql_fetch_* class of functions. Alternatively, you can use mysql_num_rows() to determine the number of rows returned in the result set.

In this case it is really senseless to wrap your actual query into a subquery. Just run your select and determine the number of rows:

$email = "[email protected]";    
$result = mysql_query("SELECT email FROM accounts WHERE email='".$email . "'");
if($result) {
    $row_count = mysql_num_rows($result);
    echo $row_count;
}

Also, you should not be writing new code using mysql_* functions, as these are deprecated. I would suggest mysqli or PDO extensions instead.

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2 Comments

It is SELECT EXISTS though which returns true or false though correct?
@user3677286 Yes. It would. But why do you care about that sort of wrapper? You will either have a row count of 0 or > 0 in the result set. That is the same thing and removes use of subselect.
0

You need to do something like

while ($r = mysql_fetch_assoc($row)) 
{ 
    echo $r[0];                                 
}

after that code.

Let me know.

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