Question

我从数据库中获取字符串，然后使用字符串构建URL。我的问题是，一些字符串将具有像＆lt; ＆GT; ＆安培; {} *一般特殊字符，但字符串也可能包含字符串。如何用短划线替换空格并完全删除字符串中的特殊字符？

Answer 1

使用preg_replace

在字符串中仅保留字母和数字

$string = preg_replace('/[^a-zA-Z0-9-]/', '', $string);

您可以使用str_replace将空格替换为 -

$string = str_replace (" ", "-", $string);

请看以下文章：

How To Clean Special Characters From PHP String

Answer 2

使用str_replace：

$str = str_replace(array(' ', '<', '>', '&', '{', '}', '*'), array('-'), $str);

注意：

如果 replace 的值少于 search ，则其余的替换值将使用空字符串。

Answer 3

1）用iconv替换变音符号 2）用空字符串替换非字母字符
3）用划线灯替换空格
4）修剪短划线字符串（也可以在操作前修剪字符串）

示例，如果您使用UTF-8编码：

setlocale(LC_ALL, 'fr_CA.utf8');
$str = "%#dŝdèàâ.,d s#$4.sèdf;21df";

$str = iconv("UTF-8", "ASCII//TRANSLIT", $str); // "%#dsdeaa.,d s#$4.sedf;21df"
$str = preg_replace("`[^\w]+`", "", $str); // "dsdeaad s4sedf21df"
$str = str_replace(" ", "-", $str); // "dsdeaad-s4sedf21df"
$str = trim($str, '-'); // "dsdeaad-s4sedf21df"

Answer 4

$search_value =array(",",".",'"',"'","\\"," ","/","&","[","]","(",")"," 
{","}",":","`","!","@","#","%","=","+");
$replace_value =array("-","-","-","-","-","-","-","-","-","-","-","-","-","-","-","- 
","-","-","-","-","-");
$product_name = str_replace($search_value,$replace_value,$row["Product_Name"]);
$product_name = str_replace("--","-",$product_name);
$product_name = str_replace("--","-",$product_name);
$product_name = preg_replace('/-$/', '', $product_name);
$product_name = preg_replace('/^-/', '', $product_name);

这将创建虚线字符串（仅包含带有破折号的字母数字字符）。可用于创建URI字符串。

Answer 5

str_replace(' ','-',$string);

字母

$output = preg_replace("/[^A-Za-z0-9]/","",$input);

如果你想保留字符：

 htmlspecialchars($string);

PHP - 字符串操作删除特殊字符并替换空格

5 个答案: