Undefined variable notice in PHP code

StackOverflow https://stackoverflow.com/questions/7625784

  •  06-02-2021
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Question

I'm getting

"Undefined variable: html in on line $html .= generateOption($optstyle.'option', $level, $data, $padding);"

What's wrong with my function? 

function generateOptions($parent, $level, $padding, $menu, $db)
{
    $result=$db->query("SELECT id, name FROM menu WHERE parent='$parent' AND showinmenu='$menu'");
    $spacer = '  '; 
    $padding = str_repeat($spacer, $level);
    while($data=$result->fetch_row()){      
        $children_html = generateOptions($data[0], $level+1, $padding, $menu,$db);
        $optstyle = empty($children_html) ? 'std' : 'bold';
        $html .= generateOption($optstyle.'option', $level, $data, $padding); (this line)
        $html .= $children_html;
    }
    return $html;
}
Was it helpful?

Solution

You didn't define $html before you tried to use it.

Try adding $html = ""; after $padding = str_repeat($spacer, $level);

OTHER TIPS

You'll have to initialise the $html before you start appending to it. Think about $html .= something like $html = $html . something and you should see the issue.

Also, your query is insecure. Make sure you escape everything - just in case.

Just declare the variable first before appending to it. Using the dot operator just appends content. If you simply use $html = ''; before your loop the warning should be gone.

function generateOptions($parent, $level, $padding, $menu, $db)
{
    $html = ''; # define `$html` first
    $result = $db->query("SELECT id, name FROM menu WHERE parent='$parent' AND showinmenu='$menu'");
    $spacer = '  '; 
    $padding = str_repeat($spacer, $level);

    while($data = $result->fetch_row())
    {      
        $children_html = generateOptions($data[0], $level+1, $padding, $menu,$db);
        $optstyle = empty($children_html) ? 'std' : 'bold';
        $html .= generateOption($optstyle.'option', $level, $data, $padding);
        $html .= $children_html;
    }
    return $html;
}

As mentioned by others, you should escape your $parent and $menu variables to prevent SQL injection.

begin the code with this : $html = "";

in begin line insert code 

if (!isset($html)) {
   $html = '';
};

        $html .= generateOption($optstyle.'option', $level, $data, $padding); (this line)
        $html .= $children_html;

Remove the dot in front of the first = in the first line.

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