在PHP的句子中找到一个单词

Question

I have this sentence:

$newalt="My name is Marie";

I want to include a page if marie or josh is found:

$words = array("marie", "josh");

$url_string = explode(" ", $newalt);

if(!in_array($url_string, $words)){
    echo "marie or josh are not in array";
}

the problem is, when I run this php it shows "marie or josh are not in array" and marie is the array (or should be). What is wrong?

Answer 1

The solution using array_intersect and array_map functions:

$newalt = "My name is Marie";
$words = ["marie", "josh"];

$url_string = explode(" ", $newalt);

if (empty(array_intersect($words, array_map("strtolower", $url_string)))) {
    echo "marie or josh are not in array";
}

http://php.net/manual/en/function.array-intersect.php

Answer 2

您在数组中错误地使用了in_array（）来检查第一个参数在第二个参数中。

in_array(strtolower($words[0]), array_map('strtolower', $url_string)) && in_array(strtolower($words[1]), array_map('strtolower', $url_string))

Answer 3

You have 2 errors. First, in_array() is case-sensitive. You need to make the haystack all lowercase first. Second, in_array() does not accept an array as the needle. To overcome this, you can use something like array_diff() :

$newalt="My name is Marie";
$words = array("marie", "josh");

$url_string = explode(" ", strtolower($newalt));

if(count(array_diff($words, $url_string) == count($words)){
    echo "marie or josh are not in array";
}

Answer 4

The first parameter of the in_array function is a array. Your $words array is filled with strings and not arrays.

Maybe try something like this:

$words = array("marie", "josh");

$urlStrings = explode(" ", $newalt);

foreach ($urlStrings as $urlString) {
    if(!in_array($urlString, $words)){
        echo "marie or josh are not in array";
    }
}