php - 从字符串中删除最后一个字符

Question

我有一个国家/地区列表作为一个数组..我希望这个数组以下列格式稍后通过 ajax 返回它：-

“印度”、“美国”、“英国” ..

使用以下代码来获得我一直在寻找的东西..

foreach($country_list as $country) {
   $countries .= '"'.$country['label'].'",';
}

问题是它给出的输出像 "india","usa","uk", ... 即带有尾随逗号。

试图删除它

substr_replace($countries, "0", -1);

和

rtrim($countries, ",");

但是没有用！.. 请帮忙！

score 7 · Accepted Answer

我认为您缺少在修剪后重新分配变量：

$s = '"india","usa","uk",';
$s = rtrim($s, ',');
// prints "india","usa","uk"
print $s;

演示

score 6 · Accepted Answer

试试这个substr()或mb_substr()

substr($string, 0, -1);
mb_substr($string, 0, -1);

或检查此链接

score 1 · Accepted Answer

您是否尝试过使用：str_replace(",", " ", $countries);

此函数应该用空格替换每次出现的逗号。

score 1 · Accepted Answer

1

if (strlen($b) > 1){
   $b = substr($b,0, -1);
}

于 2019-01-23T15:52:30.177 回答

score 0 · Accepted Answer

改用implode：

$arr = array();
foreach($country_list as $country)
  array_push($arr, $country['label']);
$comma_separated = implode(",", $arr);

score 0 · Accepted Answer

尝试这个

$countries = [];
foreach($country_list as $country) {
   $countries[] = $country['label'];
}
$new_array = implode(",", $countries);

score 0 · Accepted Answer

尝试这个

这将在每个附加字符串的开头添加一个逗号，第一个字符串除外。the 不再是要处理的尾随逗号，因此无需尝试将其修剪掉。

score 0 · Accepted Answer

// 尝试在php.ini中使用chop()函数。它有助于从字符串中删除最后一个字符

<?php
    echo '<br>' .$country_name = "'USA','UK','India',";
    echo '<br>' .chop($country_name,",");
    exit;
?>

输出：“美国”、“英国”、“印度”

score -1 · Accepted Answer

foreach($country_list as $country) {
   $countries[] = $country['label'];
}
json_encode($countries);

9 回答 9