php PHP使用preg_replace：“分隔符不能是字母数字或反斜杠”错误
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PHP using preg_replace : "Delimiter must not be alphanumeric or backslash" error
提问by Christopher
I am trying to take a string of text like so:
我正在尝试像这样获取一串文本：
$string = "This (1) is (2) my (3) example (4) text";
In every instance where there is a positive integer inside of parentheses, I'd like to replace that with simply the integer itself.
在括号内有正整数的每种情况下，我想用简单的整数本身替换它。
The code I'm using now is:
我现在使用的代码是：
$result = preg_replace("\((\d+)\)", "$result = preg_replace("((\d+))", "$result = preg_replace("#\((\d+)\)#", "", $string);
", $string);
", $string);
But I keep getting a
但我不断收到
Delimiter must not be alphanumeric or backslash.
分隔符不能是字母数字或反斜杠。
Warning
警告
Any thoughts? I know there are other questions on here that sort of answer the question, but my knowledge of regex is not enough to switch it over to this example.
有什么想法吗？我知道这里还有其他问题可以回答这个问题，但我对正则表达式的了解不足以将其切换到这个例子。
回答by codaddict
You are almost there. You are using:
你快到了。您正在使用：
$result = preg_replace("/\((\d+)\)/", "\1", $string);
- The regex you specify as the 1st
argument to
preg_*family of function should be delimited in pair of delimiters. Since you are not using any delimiters you get that error. (and)are meta char in a regex, meaning they have special meaning. Since you want to match literal open parenthesis and close parenthesis, you need to escape them using a\. Anything following\is treated literally.- You can capturing the integer
correctly using
\d+. But the captured integer will be in$1and not$0.$0will have the entire match, that is integer within parenthesis.
- 您指定为
preg_*函数族的第一个参数的正则表达式应以分隔符对分隔。由于您没有使用任何分隔符，因此您会收到该错误。 (并且)是正则表达式中的元字符，这意味着它们具有特殊含义。由于您想匹配文字左括号和右括号，您需要使用\. 以下任何内容\均按字面处理。- 您可以使用
\d+. 但捕获的整数将在$1而不是$0。$0将有整个匹配，即括号内的整数。
If you do all the above changes you'll get:
如果您进行上述所有更改，您将获得：
<?php
$string = "This (1) is (2) my (3) example (4) text";
$output = preg_replace('/\((\d)\)/i', '', $string);
echo $output;
?>
回答by animuson
1) You need to have a delimiter, the /works fine.
1）你需要有一个分隔符，/工作正常。
2) You have to escape the (and )characters so it doesn't think it's another grouping.
2）你必须转义(和)字符，所以它不认为这是另一个分组。
3) Also, the replace variables here start at 1, not 0 (0 contains the FULL text match, which would include the parentheses).
3) 此外，这里的替换变量从 1 开始，而不是 0（0 包含完整文本匹配，其中包括括号）。
$yourString='hi there, good friend';
$dividorString='there';
$someSstring=preg_replace("/$dividorString/",'', $yourString);
echo($someSstring);
// hi, good friend
Something like this should work. Any further questions, go to PHP's preg_replace()documentation- it really is good.
像这样的事情应该有效。任何进一步的问题，请转到PHP 的preg_replace()文档- 这真的很好。
回答by pix0r
Check the docs - you need to use a delimiter before and after your pattern: "/\((\d+)\)/"
检查文档 - 您需要在模式前后使用分隔符： "/\((\d+)\)/"
You'll also want to escape the outer parentheses above as they are literals, not a nested matching group.
您还需要转义上面的外括号，因为它们是文字，而不是嵌套的匹配组。
请参阅：preg_replace 手册页
回答by Atli
Try:
尝试：
##代码##The parenthesis chars are special chars in a regular expression. You need to escape them to use them.
括号字符是正则表达式中的特殊字符。你需要逃避它们才能使用它们。
回答by Darian Harrison
Delimiter must not be alphanumeric or backslash.,
分隔符不能是字母数字或反斜杠。,
try typing your parameters inside "/ .... /" as shown bellow. Else the code will output >>> Delimiter must not be alphanumeric or backslash.
尝试在“/ .... /”中输入参数，如下所示。否则代码将输出 >>> 分隔符不能是字母数字或反斜杠。
##代码##. . worked for me.
. . 为我工作。
