Define a variable in a function php [duplicate] - php

Question

This question already has an answer here:
Notice : Undefined variable when concatenating
(1 answer)
Closed 7 years ago.
I have a function as below but I get warning for $tab variable that says it is not defined. How can I define it and not receive this warning anymore?
<?php
/*** suggested articles as random ***/
function doArticle_suggested_small_horizontall($articleid,$title,$photo,$parentid,$catid,$altdescription) {
$tab .= "<table width=150 cellspacing=5 style=border: 1px solid #0066ff align=right>\n";
$tab .= "<tr>\n";
$tab .= "<td align=center bgcolor=#ffffff><a href='../artandculture/adetails.php?articleid=$articleid&parentid=$parentid&catid=$catid'>
<img src='../images/simage/$photo' border='0' alt='$altdescription'></a></td>\n";
$tab .= "</tr>\n";
$tab .= "<tr align=right width=150 height=80 border=0 style=border: 1px solid #ffffff>\n";
$tab .= "<td width=110 align=right dir='rtl' border=0 style=border: 1px solid #ffffff ><p class=articletitlenounderline><a href='../artandculture/adetails.php?articleid=$articleid&parentid=$parentid&catid=$catid'><strong>$title </strong></p></a></td>\n";
$tab .= "</tr>\n";
$tab .= "</table> <p> <hr class='hr99' ></hr></p>";
return $tab;
}
$tab = "";
?>

Answer 1

You need to define $tab at the beginning of the function.
replace $tab .= "<table width=15 ....
with $tab = "<table width=15....
Or you could just add $tab = ""; as the first line of the function, now you are defining it outside the function which you should remove.

Answer 2

Define $tab,
$tab = [...];
Then you need to do .= on the variable. The variable does not exist where you start using it.

Answer 3

".=" is telling PHP to add a string to an existing variable. PHP cannot add the string if there is no existing variable.
// Add this before your first $tab .=
$tab = NULL;

Answer 4

Move $tab to the start of the function in order for the concatenation to properly work:
Instead of this in the end:
$tab = "";
Do this:
<?php
/////////////////////suggested articles as random /////////////////////////
function doArticle_suggested_small_horizontall($articleid,$title,$photo,$parentid,$catid,$altdescription) {
$tab = "";
$tab .= "<table width=150 cellspacing=5 style=border: 1px solid #0066ff align=right>\n";
$tab .= "<tr>\n";
$tab .= "<td align=center bgcolor=#ffffff><a href='../artandculture/adetails.php?articleid=$articleid&parentid=$parentid&catid=$catid'>
<img src='../images/simage/$photo' border='0' alt='$altdescription'></a></td>\n";
$tab .= "</tr>\n";
$tab .= "<tr align=right width=150 height=80 border=0 style=border: 1px solid #ffffff>\n";
$tab .= "<td width=110 align=right dir='rtl' border=0 style=border: 1px solid #ffffff ><p class=articletitlenounderline><a href='../artandculture/adetails.php?articleid=$articleid&parentid=$parentid&catid=$catid'><strong>$title </strong></p></a></td>\n";
$tab .= "</tr>\n";
$tab .= "</table> <p> <hr class='hr99' ></hr></p>";
return $tab;
}
?>

Answer 5

In addition to the other answers, if you already have something assigned to $tab and want to append it from function, you must either send it in as an argument to the function or add the function output
function example( $tab, $something = false ) {
$tab .= ', added this in example function';
return $tab;
}
$tab = 'Original content';
echo example( $tab ); // Output: Original content, added this in example function
function example2( $something = false ) {
return ', added this in example2 function';
}
$tab = 'Original content 2';
$tab .= example2();
echo $tab: // Output: Original content 2, added this in example2 function