Foreach在php中不起作用

Question

I have a code like this:

<?php

$kode ["J"]= array (20, C, D, F);
$kode ["K"]= array (50, B, G, U);
$kode ["T"]= array (70, V, W);

function kota ($start, $end){
    if (is_array($kode)) {
        foreach ($kode as $kota => $path){
            if ($kota=$end) {
                for ($i=1; $i < count ($kota); $i++){
                    $jalur=$start.$path[$i];
                }
            }
        }
        return $jalur;
    }
}
$start = "J";
$end = "T";
$hasil=kota ($start, $end);
echo "".$hasil;
?>

I want the output to be JVW
I don't know what is wrong, can anyone help me? please...

Answer 1

Syntatical errors

Looks like you forgot to use the equality operator ==

if ($kota = $end){ ... }

Should be -

if ($kota == $end){ ... }

By using only one equals sign you are actually assigning a value to $kota , not comparing the value to $end as should be done in conditional expressions.

I don't think this is the only thing that is causing trouble here.. but it definitely should be sorted out :)

---

Variable scope

Another thing I noticed in your code is that you are referencing variables within the kota function that were not defined in it's scope. This means that the $kota array is not accessible within the kota function. You should pass the $kota array to the function so that you can use it within scope of the function. Here is some more info on variable scopes in PHP .

---

Naming conventions

One final note on your variable name choice... You should possibly think of changing the variable $kota or function kota so that their names are not identical. This will help improve readability and perhaps prevent some mistakes at 4am when you've been debugging the whole night ;)

Answer 2

On the line

if ($kota=$end){

you are not comparing, but overwriting the value in $kota, and that is always true.

Also the $kode is not available in the function scope, try adding it to the parameter list, or using global (not advised).

Answer 3

You need to either pass $kode into your function as an argument or call global $kode; inside your function. I'd recommend the former.

Additionally, if ($kota=$end) needs to be if ($kota==$end) as others have mentioned.

Answer 4

Not sure, that $jalur=$start; is on the correct place, but this script gives what you want:

<?php

$kode ["J"]= array (20, 'C', 'D', 'F');
$kode ["K"]= array (50, 'B', 'G', 'U');
$kode ["T"]= array (70, 'V', 'W');

function kota ($start, $end){
      global $kode;

      if(is_array($kode)){

            foreach ($kode as $kota => $path){

                  if ($kota == $end){
                        $jalur=$start;
                        for ($i=1; $i < count ($path); $i++){
                              $jalur .= "-" . $path[$i];
                        }
                  }
            }
            return $jalur;
      }
}
$start  = "J";
$end    = "T";
$hasil=kota ($start, $end);
echo $hasil;
?>

Answer 5

Your code has lots of issues

1. As others pointed out, in if ($kota=$end) { , you are assigning $end to $kota and always return true ie the code in the IF clause always execute

2. PHP has function scope. Simply put, variables declared inside a function cannot be used outside, and vice versa. Use parameters to pass your $kode into the function.

3. Use of bare strings ie V and W in $kode ["T"]= array (70, V, W); and other places. This is highly recommended against, and PHP does warn you about this.

4. As other pointed out, $jalur=$start.$path[$i]; would overwrite $jalur every time. The for-loop outside is meaningless. You would use .= the append-to operator. Note that you also need to initialize your variable before using this operator.

5. $kota is always a string in your code, because in a foreach loop, the variable before => symbol means get the key of the array, and array keys can only be either String or integer. That said, for ($i=1; $i < count ($kota); $i++){ is meaningless because count($kota) cannot be greater than 1 - your for loop actually never runs.

6. This is blatantly meaningless to append a variable with an empty string as in echo "".$hasil;

I guess this is what you want.

<?php

$kode ["J"] = array (20, 'C', 'D', 'F');
$kode ["K"] = array (50, 'B', 'G', 'U');
$kode ["T"] = array (70, 'V', 'W');

function kota ($kode, $start, $end){
    $jalur = $start;
    if (is_array($kode)) {
        foreach ($kode as $kota => $path){
            if ($kota == $end) {
                for ($i = 1; $i < count($path); ++$i) {
                    $jalur .= '-' . $path[$i];
                }
            }
        }
        return $jalur;
    }
}
$start = "J";
$end = "T";
$hasil = kota($kode, $start, $end);
echo $hasil;
?>

This code give you the string which starts with $start and all other elements except the first element in $kode[$end]