Trying to construct a regex that will locate a pattern of ANY character followed by double quotes
This regex locates each occurrence properly
(\S"")
Given the example below
$string='"WEINSTEIN","ANTONIA \"TOBY"","STILES","HOOPER \"PETER"","HENDERSON",';
$pattern = '(\S"")';
$replacement = '\\""';
$result=preg_replace($pattern, $replacement, $string);
My result turns out to be
"WEINSTEIN","ANTONIA \"TOB\"","STILES","HOOPER \"PETE\"","HENDERSON"
But I am seeking
"WEINSTEIN","ANTONIA \"TOBY\"","STILES","HOOPER \"PETER\"","HENDERSON"
I understand the replacement is removing/replacing the whole match, but how can I remove all but the first letter rather than completely replacing it?
You can change your pattern to use a positive lookbehind instead so that it doesn't capture the non-space character:
$string='"WEINSTEIN","ANTONIA \"TOBY"","STILES","HOOPER \"PETER"","HENDERSON",';
$pattern = '/(?<=\S)""/';
$replacement = '\\""';
$result=preg_replace($pattern, $replacement, $string);
echo $result;
Output
"WEINSTEIN","ANTONIA \"TOBY\"","STILES","HOOPER \"PETER\"","HENDERSON",
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