CODEWITHSUNDEEP
phpmysqlif-statement
DMSJax
DMSJax

Reputation: 1727

Broken If statment in SQL Query

Ok Im shaking my head at myself as I know this is something obvious but im just not getting it. The query should return 1 result with a matching username and password, then - i just need to check if the column/field "Status" in the row is set to Inactive or Suspended -- If yes then set value of $count and set a $_SESSION value of msg.

As it stands right now the check of Status just doesn't work - Is it because of how I'm comparing the column.field name? $result['Status']?

my query is as follows:

$sql="SELECT * FROM $tbl_name WHERE User_Name='$myusername' and Password='$mypassword'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);

// Verified E-Mail address check
if ($result['Status']=="Inactive") {$count=5;$_SESSION['msg']=="Not Verified";};

Upvotes: 0

Views: 148

Answers (2)

Goran Lepur
Goran Lepur

Reputation: 594

You should use mysql_fetch_array or mysql_fetch_assoc to get an array of data from your query.

$sql="SELECT * FROM $tbl_name WHERE User_Name='$myusername' and Password='$mypassword'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);

// Verified E-Mail address check

$row = mysql_fetch_array($result);

if ($row['Status']=="Inactive") {$count=5;$_SESSION['msg']=="Not Verified";};

Also, my advice is not to use mysql_ functions at all as they are deprecated from version 5.5. Check out mysqli or even better pdo

Upvotes: 3

jcsanyi
jcsanyi

Reputation: 8174

You need to fetch your data row from the result before you can compare values.

Something like this:

$result = mysql_query($sql);
$data = mysql_fetch_array($result);
if ($data['Status'] == 'Inactive') {
    ...
}

That being said, there's good reasons not to use mysql_* functions.
See this question for more details: Why shouldn't I use mysql_* functions in PHP?

Upvotes: 3

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