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PHP - String Replace SyntaxError: illegal character

Output for $status

Array
(
    [1] => 1
    [2] => 0
    [3] => 0
    [4] => 4
    [5] => 4
)

$color_code_string = implode(",",$status);

Ouput

1,0,0,4,4

$color_code_string = str_replace("0","'#F00'",$color_code_string); 
$color_code_string = str_replace("1","'#00bcd4'",$color_code_string);
$color_code_string = str_replace("2","'#4caf50'",$color_code_string);
$color_code_string = str_replace("3","'#bdbdbd'",$color_code_string);
$color_code_string = str_replace("4","'#ff9900'",$color_code_string);

Exception

SyntaxError: illegal character
colors: ['#00bcd'#ff9900'','#F00','#F00','#ff9900','#ff9900']

//prints '#00bcd'#ff9900'','#F00','#F00','#ff9900','#ff9900'

How do I achieve expected output as below

'#00bcd','#ff9900','#F00','#F00','#ff9900','#ff9900'
like image 823
Slimshadddyyy Avatar asked Jan 26 '26 17:01

Slimshadddyyy


2 Answers

That happens because you are also replacing numbers inside the color codes you replaced before. Solution: traverse the array to do the replacement before imploding the array of colors:

// Translation table, saves you separate lines of stringreplace calls.
$colorCodes = array(
  0 => "#F00",
  1 => "#00bcd4",
  2 => "#4caf50",
  3 => "#bdbdbd",
  4 => "#ff9900",
);

// Build an array of colors based on the array of status codes and the translation table.
// I'm adding the quotes here too, but that's up to you.
$statusColors = array();
foreach($status as $colorCode) {
  $statusColors[] = "'{$colorCodes[$colorCode]}'";
} 

// Last step: implode the array of colors.
$colors = implode(','$statusColors);
like image 121
GolezTrol Avatar answered Jan 28 '26 07:01

GolezTrol


$status = [1,0,0,4,4,];
$color_code_string = implode(",",$status);
$replacements = ["0" => "'#F00'","1" => "'#00bcd4'","2" => "'#4caf50'","3" => "'#bdbdbd'","4" => "'#ff9900'",];
$color_code_string = strtr($color_code_string, $replacements); 
echo $color_code_string;
like image 35
Amit Avatar answered Jan 28 '26 07:01

Amit



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