i want to use the variable out of the loop but the variable defined in the while loop

Question

Shows "Undefined variable: display1". I am able to display the while loop. The problem is $display1 had define in while loop and it must in it because have $i increment.

//SQL for display category name
$query1="SELECT cat_name FROM restaurant_category WHERE rest_id={$display}";
$result1=mysql_query($query1);
$rowNum1=mysql_num_rows($result1);
//SQL for display dish information of each category
$query3="SELECT cat_id FROM restaurant_category WHERE cat_name={$display1}";
$result3=mysql_query($query3);
$display3=mysql_query($result3);
$query2="SELECT dish_name, dish_description, dish_price FROM dish WHERE cat_id={$display3}";
$result2=mysql_query($query2);
$rowNum2=mysql_num_rows($result2);
$data=mysql_fetch_row($result2);

while($i<$rowNum1)
{
$display1=mysql_result($result1,$i,"cat_name");
//display category name
    while($j<$rowNum2)
    {//display dish information}
}

PHP is not going to time travel for you and execute the while loop to fetch results BEFORE you've actually run the query to PRODUCE the results in the first place. — Marc B
– Marc B, Commented Jul 4, 2013 at 19:32

beiller · Accepted Answer · 2013-07-04 19:30:28Z

1

//SQL for display dish information of each category
$query3="SELECT cat_id FROM restaurant_category WHERE cat_name={$display1}";
$result3=mysql_query($query3);
$display3=mysql_query($result3);

This appears to be an error, calling mysql_query on $result3. It should probably be

$display3 = mysql_fetch_row($result3)

And here:

$display1 = null;
while($i<$rowNum1)
{
$display1=mysql_result($result1,$i,"cat_name");
//display category name
    while($j<$rowNum2)
    {
       //display dish information
    }
}

Possible syntax errors in there. Also add $display1 outside of the while scope.

Also oblig: mysql_* functions are depreciated.

answered Jul 4, 2013 at 19:30

beiller

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3 Comments

Jon Over a year ago

Declaring $display1 outside makes no difference in PHP. They are not using it anywhere -- at least in the code shown.

Marc B Over a year ago

OP wants to time travel and use $display1 before it's ever been assigned a value.

beiller Over a year ago

If you actually read the code, they assign $display1 a value. That's what a "=" sign means in PHP. Assignment. See comments such as "display category name" probably has something to do with $display1 variable after its assignment.

Vegard Larsen · Accepted Answer · 2013-07-04 19:31:26Z

0

The $display1 variable will be defined after the while loop. The warning you are seeing is because you are assigning to a variable you haven't used before inside a loop. Simply declare the variable outside the while loop to get rid of the warning:

$display1 = null;
while ($i<...

answered Jul 4, 2013 at 19:31

Vegard Larsen

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