Question

我正面临一种情况，如果地址长于80个字符，我需要将address分为两部分，即address1和address2。

我可以在strlen下添加以下条件，以查找地址是否大于80个字符，但是当地址长于80个字符时，在逗号后将地址平均分割会遇到问题。

我的代码如下：-

// Sample Address
$address = "House No. 1234-ABC, Second Floor, XYZ Building, ABC Mall Road, Street Number 1234, ABC Area";

$address_length = strlen($address);

if(address_length > 80){
    // Split Long Address Code
    $address_equal_length = $address_length/2;
    $address1 = substr($address,0,$address_equal_length);
    $address2 = substr($address,$address_equal_length,$address_length); 
}

不幸的是，上面的代码将不会检查逗号并按如下所示拆分地址：-

$address1 = "House No. 1234-ABC, Second Floor, XYZ Buildin";
$address2 = "g, ABC Mall Road, Street Number 1234, ABC Area";

您可以看到它剪切了Building字并在Buildin的末尾显示为$address1，在g的开头显示为$address2

我希望$address如下划分：-

$address1 = "House No. 1234-ABC, Second Floor, XYZ Building";
$address2 = "ABC Mall Road, Street Number 1234, ABC Area";

希望有人可以帮助我！在此先感谢：）

Answer 1

一个好的开始可能是：

// Sample Address
$address = "House No. 1234-ABC, Second Floor, XYZ Building, ABC Mall Road, Street Number 1234, ABC Area";

$address_length = strlen($address);

if($address_length > 80){
    // Split Long Address Code
    $address_equal_length = $address_length/2;
    $commaIndex = strrpos(substr($address,0,$address_equal_length),',');
    $address1 = substr($address, 0,$commaIndex);
    $address2 = substr($address,$commaIndex + 1); 
}

您需要测试$ commaIndex是否存在，以确保substr不会失败。

修改

一种优化的解决方案，它在字符串的中心寻找最接近的逗号：

<?php

// Sample Address
$address = "House No. 1234-ABC, Second Floor, XYZ Building, ABC Mall Road, Street Number 1234, ABC Area";

$address_length = strlen($address);

if($address_length > 80){
    // We gonna look for the comma which is the more close to the center of the string
    $address_equal_length = $address_length/2;
    $comma_index_before = strrpos(substr($address, 0, $address_equal_length), ',');
    $comma_index_after = strpos($address, ',', $address_equal_length);
    if ($comma_index_before === false) {
        $comma_index_before = 0;
    }
    if ($comma_index_after === false) {
        $comma_index_after = $address_length;
    }
    $comma_index = $address_equal_length - $comma_index_before <= $comma_index_after - $address_equal_length ? $comma_index_before : $comma_index_after;

    $address1 = substr($address, 0, $comma_index);
    $address2 = substr($address, $comma_index + 1); 
}

Answer 2

动态正则表达式是最直接/最适合此任务的工具。

代码：（Demo）

var_export($address1);
echo "\n";
var_export($address2);

在我的演示中，输出来自：

'House No. 1234-ABC, Second Floor, XYZ Building'
'ABC Mall Road, Street Number 1234, ABC Area'

是：

var_export()

我正在使用.{{$middle}}来显示值，以证明该解决方案在分割点处包括空白修剪。

模式分类为：

.{45}的评估结果为[^,]*，表示匹配45个出现的任意字符
\K匹配零个或多个非逗号字符
, ?重新开始全串匹配，也就是忘记先前匹配的字符
preg_split()匹配要分割的逗号和后面的空格（可选）

我在2的{{1}}上设置了硬性限制，只是为了明确/安全。

我以前在展示放置好的省略号时曾推荐这种技术（了解更多）：How to add an ellipsis hyperlink after the first space beyond 170 characters?

Answer 3

如果地址总是被2个逗号分隔，则可以使用explode，例如array_chunk来获取多个项目。然后使用implode重新组装零件。

这还将适用于2个以上的地址。

$address = "House No. 1234-ABC, Second Floor, XYZ Building, ABC Mall Road, Street Number 1234, ABC Area";
$address_length = strlen($address);

if ($address_length > 80) {
    $parts = explode(',', $address);
    list($address1, $address2) = array_map(function($x) {
        return trim(implode(',', $x));
    }, array_chunk($parts, 3));
}

Php demo

Answer 4

这是我的主意首先，您将所有逗号部分放入一个数组中，然后对其进行迭代，以查看它们加到前一个长度上是否超过80（如果您转到下一行）

   $addressParts=explode(",",$address);
   $endString="";
   $currentLine="";
   foreach($addressParts as $addressPart){
       if(str_len($currentLine.$addressPart)>80){
           $endString=$endString.","$currentLine."<br>";
           $currentLine=$addressPart;
       }else{
           if($currentLine != ""){
              $currentLine=$currentLine.", ".$addressPart;
           }else{
              $currentLine=$currentLine.$addressPart;
            }
       }
   }
   $endString=$endString.$currentLine;

   echo $endString;

在PHP中，逗号后将地址平均分为两部分

4 个答案: