PHP uses a foreach loop to iterate through the array, and the loop executes successfully, but the loop is not successful? -Segmentfault

            foreach($arr as $value){                $value=preg_replace("/href=\"\//i",'href="'.$link,$value);                echo $value."
";//此处$value为替换之后的值。            }            echo $arr[1];//此处显示的仍为替换前$arr[1]的值。//请问这个是怎么回事呢？``

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            foreach($arr as $value){                $value=preg_replace("/href=\"\//i",'href="'.$link,$value);                echo $value."
";//此处$value为替换之后的值。            }            echo $arr[1];//此处显示的仍为替换前$arr[1]的值。//请问这个是怎么回事呢？``

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4 answers

The answer is helpful to the person, has the reference value 0 The answer is no help, it's the wrong answer, irrelevantly replying .

foreach($arr as &$value){    $value=preg_replace("/href=\"\//i",'href="'.$link,$value);    echo $value."
";//此处$value为替换之后的值。}echo $arr[1];//此处显示的仍为替换前$arr[1]的值。

Because of the scope of the variable, if you modify the value inside, and you need to save the result, you need to define the type as a reference type , that is & .

Change it.

        foreach($arr as $key=>$value){            $arr[$key]=preg_replace("/href=\"\//i",'href="'.$link,$value);            echo $arr[$key]."
";//此处$value为替换之后的值。        }        

You just did a replace operation on the local variable $value in the loop, and did not return the value to $arr.

foreach($arr as $key => $value){    $value=preg_replace("/href=\"\//i",'href="'.$link,$value);    $arr[$key] = $value;    echo $value."
";//此处$value为替换之后的值。}echo $arr[1];//此处显示的仍为替换前$arr[1]的值。

Get it straight, pass it by value, and pass it by reference.

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