Question

我有一个从输入字段生成的字符串，我想检查前两个字符，看看它们是否在数组中找到。如果他们是我想要一条消息出现。

任何人都可以解释为什么这不起作用吗？

$i = strtoupper($_POST['postcode']);
    $ep = array("AB", "BT", "GY", "HS", "IM", "IV", "JE", "PH", "KW");

    if (isset($i)) {

    if(substr($i, 0, 2) === in_array($i, $ep)) {
        echo "Sorry we don't deliver to your postcode";
    }   
}

Answer 1

您对in_array的使用是错误的。改变

if(substr($i, 0, 2) === in_array($i, $ep)) {
  echo "Sorry we don't deliver to your postcode";
}

到

if(in_array(substr($i, 0, 2), $ep)) {
  echo "Sorry we don't deliver to your postcode";
}

Answer 2

您误解了in_array的工作原理，请查看manual了解详情。

以下代码是检查给定邮政编码是否有效的改进方法

<?php
/**
 * Check if the given post code is valid
 * @param string $postcode 
 * @return boolean
 */
function is_valid_postcode( $postcode = '' )
{
    $ep = array("AB", "BT", "GY", "HS", "IM", "IV", "JE", "PH", "KW");
    $postcode = strtoupper( $postcode );
    return in_array( $postcode , $ep );
}

if( isset( $_POST['postcode'] ) ){

    // Remove unwanted spaces if they're there
    $postcode = trim( $_POST['postcode'] );

    // Extract only the first two caracters
    $postcode = substr($postcode, 0, 2 );

    // Check if the submitted post code is valid
    if( !is_valid_postcode( $postcode ) ){
        echo "Sorry we don't deliver to your postcode";
    }
}

Answer 3

像这样使用：

$i = strtoupper($_POST['postcode']);
$ep = array("AB", "BT", "GY", "HS", "IM", "IV", "JE", "PH", "KW");

if (isset($i)) {

    $i = substr($i, 0, 2);

    if(in_array($i, $ep)) {
        echo "Sorry we don't deliver to your postcode";
    }   
}

Answer 4

试试这个：

if(in_array(substr($i, 0, 2), $ep)) {
    echo "Sorry we don't deliver to your postcode";
}

Answer 5

试试这样：

$i = strtoupper($_POST['postcode']);
    $ep = array("AB", "BT", "GY", "HS", "IM", "IV", "JE", "PH", "KW");

    if (isset($i)) {
    $i = substr($i, 0, 2);
    if(in_array($i, $ep)) {
        echo "Sorry we don't deliver to your postcode";
    }   
}

substr前两个字母匹配数组不起作用

5 个答案: