Question

我有字符串$text = " /var/www/images/mobile/test/test/, 1346744549";，我需要将其转换为$text = " '/var/www/images/mobile/test/test/', '1346744549'"; - 将'添加到字符串中的每个“值”。问题出现斜杠，我不知道如何识别它。这是我的样本，但它已经错了......

$text = " /var/www/images/mobile/test/test/, 1346744549";
$text = preg_replace("/\b|\/\b/i", '"', $text);
echo $text;

Answer 1

这是功能：

"'".implode("','",explode(',',$text))."'";

您可以在此处查看结果：http://sandbox.onlinephpfunctions.com/code/14ed966d086494933f0e0ff48230083623a9c527

Answer 2

尝试

$text = " /var/www/images/mobile/test/test/, 1346744549";
$text = preg_replace("/[^\s,]+/", "'$0'", $text);
echo $text;

[^\s,]+匹配任何系列的非空格，非逗号字符本身，但周围有'（$ 0是匹配）

如果您希望允许数据中的空格，请尝试使用

$text = " /var/www/images/mobile/test/test/, 1346744549, Hello Foobar test";
$text = preg_replace("/(^\s*|,\s*)([^,]+)/", "$1'$2'", $text);
echo $text;

将输出

'/ var / www / images / mobile / test / test /'，'1346744549'，'Hello Foobar test'

Answer 3

$text = " /var/www/images/mobile/test/test/, 1346744549";
echo preg_replace('/(?=[,\s]|$|^)/i', '"', $text);

如何正确编写RegEx

3 个答案: