CODEWITHSUNDEEP
phpsyntax
theking963
theking963

Reputation: 2216

PHP Syntax in functions

I am using a loop to go through the result of query and display the data in a grid. I only want to display the delete button in certain rows. I can setup if else statements for that.

while($rows = mysql_fetch_array($result)){

//HTML code - I am not using echo here just plain html code written below

deleteButton();//I am calling the deleteButton function here.

//HTML code

}

I have a deleteButton function at the top of the page

function deleteButton()
{
echo "<form name='DeleteInfo' method='POST' action='delete.php'>";
echo "<input type='hidden' name='val1' value=$rows[col1]>";
echo "<input type='hidden' name='val2' value=$rows[col2]>";
echo "<input class='center' type='submit' value='Delete'/>";
echo "</form>";
}

I get an error undefined variable rows for col1 and col2. I am assuming it's a syntax problem.

I have also tried.

echo "<input type='hidden' name='val2' value=$rows['col2']>";
echo "<input type='hidden' name='val2' value='$rows[col2]'>";

This HTML code works

<input type="hidden" name="name1" value="<?php echo $rows['col3']; ?>">

EDIT: Resolved. It was out of scope. I have to use global $rows in the function. Thanks Marc.

Upvotes: 1

Views: 153

Answers (5)

Paul DelRe
Paul DelRe

Reputation: 4039

As Steven pointed out, you have to pass $rows into the function as well.

Full example:

function deleteButton($row)
{
    echo "<form name='DeleteInfo' method='POST' action='delete.php'>";
    echo "<input type='hidden' name='val1' value={$rows['col1']}>";
    echo "<input type='hidden' name='val2' value={$rows['col2']}>";
    echo "<input class='center' type='submit' value='Delete'/>";
    echo "</form>";
}

while($rows = mysql_fetch_array($result)){

    // ...

    deleteButton($rows);

    // ...

}

You could wrap your array accessor in braces or use concatenation

echo "<input type='hidden' name='val2' value={$rows['col2']}>";
echo "<input type='hidden' name='val2' value=".$rows['col2'].">";

Upvotes: 1

Pheonix
Pheonix

Reputation: 6052

$rows is not accessable from that function, so what you can do is either pass it as an argument to that function, for make the $rows GLOBAL, by defining it outside any scope and putting the statement

global $rows;

in the functions where you want to access it.

Upvotes: -1

Spudley
Spudley

Reputation: 168655

Variables do not stay in scope from one function to the next. If you have a variable that you want to use in a function that you're calling, then you need to pass that variable into the function. (the other option is to make them global, but I'll ignore that, as it's not a good idea to overuse globals)

So where you call the function, you need to call it like this:

deleteButton($rows);

...and then in the function declaration, you need to add it there as well, like this:

function deleteButton($rows)

This is fairly basic stuff as far as PHP goes, or indeed programming in general. It would be good for you to get a handle on this kind of thing, make sure you understand it properly befoe you write too much more code.

Upvotes: 1

Steven Dickinson
Steven Dickinson

Reputation: 503

Your $rows variable doesn't exist in the scope of the deleteButton function. You could amend it like this:

function deleteButton($rows) {
    // etc
}

and then call it as:

deleteButton($rows);

Upvotes: 4

marchaos
marchaos

Reputation: 3444

echo "<input type='hidden' name='val2' value=".$rows['col2'].">";

Upvotes: 0

Related Questions