ctype_space

(PHP 4 >= 4.0.4, PHP 5, PHP 7, PHP 8)

ctype_space — 检测空白字符

说明

ctype_space(mixed $text): bool

检测提供的 string 类型的 text 里面的所有字符是否都是创建空白。

参数

text: 要测试的字符串。
注意:
如果给出一个 -128 到 255 之间(含)的int, 将会被解释为该值对应的ASCII字符 (负值将加上 256 以支持扩展ASCII字符). 其它整数将会被解释为该值对应的十进制字符串.

警告
自 PHP 8.1.0 起，弃用传递非字符串参数。未来该参数将解释为字符串而不是 ASCII 码点。根据预期行为，应将参数转为字符串或显式调用 chr()。

返回值

如果 text 里面的每个字符都创建某种形式的空白，那么就返回 true；否则返回 false。除了空白字符，还包括制表符、垂直制表符、换行符、回车和换页符。当使用空字符串调用时，结果始终为 false。

示例

示例 #1 ctype_space() 示例

<?php
$strings = array(
'string1' => "\n\r\t",
'string2' => "\narf12",
'string3' => '\n\r\t' // 注意单引号
);
foreach ($strings as $name => $testcase) {
    if (ctype_space($testcase)) {
        echo "The string '$name' consists of whitespace characters only.\n";
    } else {
        echo "The string '$name' contains non-whitespace characters.\n";
    }
}
?>

以上示例会输出：

The string 'string1' consists of whitespace characters only.
The string 'string2' contains non-whitespace characters.
The string 'string3' contains non-whitespace characters.

参见

ctype_cntrl() - 检测控制字符
ctype_graph() - 检测除空格外的任何打印字符
ctype_punct() - 检测可打印的字符是不是不包含空白、数字和字母
IntlChar::isspace() - Check if code point is a space character

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用户贡献的备注 2 notes

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gardnerjohng at hotmail dot com ¶

15 years ago

A function I wrote last night was fairly flexible in terms of detecting whitespace, and even took into account the pesky non-breaking spaces / zero-width spaces further up the Unicode alphabet.


The benefit here was being able to isolate and identify specific Unicode indices based on their subrange.


<?php

//    Returns TRUE if the ASCII value of $string matches a registered whitespace character.

    //    *    This includes non-breaking spaces, zero-width spaces, and any unicode values below 32.

    //    *    $string: Character to identify. If string extends past one character, the value

    //        is truncated and only the initial character is examined.

function is_whitespace($string){

//    Return FALSE if passed an empty string.

if($string == "") return FALSE;


$char    =    ord($string);


//    Control Characters

if($char < 33)                        return TRUE;

        if($char > 8191 && $char < 8208)    return TRUE;

        if($char > 8231 && $char < 8240)    return TRUE;


//    Additional Characters

switch($char){

            case 160:    // Non-Breaking Space

case 8287:    // Medium Mathematical Space

return TRUE;

                break;

        }

        return FALSE;

    }

?>

down

vlabots at gmail dot com ¶

10 years ago

thanks to gardnerjohng, but a had some problems with non-breaking spaces in this function.

I added 2 more cases for this: 
case 0xC2
case 0xA0

After this modification non-breaking spaces in my test code were successfully detected.