如何检查列表中的一个var？ 而不在列表中

Question

I have the following array.

$ar = array("apple:3","banana:2","canopule:6","grapes:5","cherry:8");

How can I check to see if the following variable

$a = "grapes";

Is in the array (in this case grapes:5) and if it is, return the matched string. I.e., "grapes : 5"

If it's not in the list (e.g., matching "strawberry"), return "strawberry : not found."

I have this:

for($i = 0; $i < count($ar); $i++) {
    list($name,$count)=explode(":",$ar[$i]);
    if($a==$name)
        echo "$a : $count";
    else
        echo "$a : not found!";
}

If it matches, it works fine. But if not, I get

strawberry : not found!strawberry : not found!strawberry : not found!strawberry : not found!strawberry : not found!

It returns every false case instead of just one.

Answer 1

Try like this:

$ar=array("apple:3","banana:2","canopule:6","grapes:5","cherry:8");
$a = "strawberry";
$result = '';
for($i=0;$i<count($ar);$i++){
    list($name,$count)=explode(":",$ar[$i]);
    if($a == $name){
        $result = "$a : $count";
        break;
    }
}

if($result){
    echo $result;
} else {
    echo $a." : Not found";
}

Simply check if there is something in loop if yes print result, if no add not found.