Session Variable in Select Statement

Question

On one page i have it to show the user who logged in username which works perfectly fine.

echo "Welcome, ".$_SESSION['username']

I wanna create a Select Statement which uses the same variable as above to pull all data for that user.

Example Username is [email protected]

I used this to see if the coding works which it did.

$sql="SELECT * FROM $tbl_name WHERE myusername ='[email protected]'";
$result=mysql_query($sql);

But I wanna use the variable; I tried the code below but it didn't work.

$sql="SELECT * FROM $tbl_name WHERE myusername ='$_SESSION['username']'";
$result=mysql_query($sql);

I know for sure it has to do with the SELECT statement using the variable, im not sure if i am relating to the variable correctly.

Help will be greatly appreciated.

Solution 1

Try using

$sql="SELECT * FROM $tbl_name WHERE myusername ='{$_SESSION['username']}'";

This syntax where you use the variable replacement capacity of PHP in string is much easier to read, at-least when you have to replace multiple variables in a string.

Solution 2

Basic PHP syntax rules: When embedding an array reference in a string, you do NOT use quotes on the keys:

echo "$arr['key']";   // wrong
echo "$arr[key]";     // right
echo "{$arr['key']}"; // right

Note that the {} syntax is REQUIRED if you're embedding a multi-dimensional array:

echo "$arr[foo][bar]";

is parsed as

echo $arr['foo'];
echo '[bar]';

Adding the {} makes PHP use the entire array key sequence:

echo "{$arr['foo']['bar']}";

Solution 3

Your returning a string of myusername = $_SESSION['username'], when you actually need whats stored in that. Remove the '' and append the variable to the string.

Something like this:

$sql= 'SELECT * FROM $tbl_name WHERE myusername = ' . $_SESSION['username'] . '';

Solution 4

$result = mysql_query("SELECT * FROM tbl_task where username =  '".$_SESSION['username']."' ORDER By DateAssigned ASC ");

Solution 5

$sql="SELECT * FROM $tbl_name WHERE username ='".$_SESSION['username']."'";
$result=mysqli_query($sql);

Try this ........

100% it will work.

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