1
Good afternoon, stackoverflow crew. Is there a date in an array? I’ve been trying, using it like this:
for($i = 0; $i < count($array); $i++){
if(preg_match( "\d{1,2}/\d{1,2}/\d{4}$" , $array[$i] )){
echo 'true';
}else{
echo 'false';
}
}
Only you’ve been making mistakes:
Delimiter must not be alphanumeric or backslash in C: wamp www ecoprintQ ecoLicenseLayout json dados-partners-r epots.php on line 14
Someone knows how to do it?
3
The delimiters are missing, the functions preg demand this and perhaps also lack the ^ in regex \d{1,2}/\d{1,2}/\d{4}$, do so #^\d{1,2}/\d{1,2}/\d{4}$#:
for($i = 0; $i < count($array); $i++){
if(preg_match( "#^\d{1,2}/\d{1,2}/\d{4}$#" , $array[$i] )){
echo 'true';
}else{
echo 'false';
}
}
Understand what delimiters are and how they work on PCRE: http://php.net/manual/en/regexp.reference.delimiters.php
Warning: preg_match(): No ending delimiter '#' found in
well, thanks, it worked out
@gabrielfalieri blz, that good!
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Who mistakes? Error of
parse? Returns something liketrueshouldn’t you? Please be clearer.– Guilherme Nascimento
Give an example of some value that fails and another that must pass.
– rray
Must fail, any string "Hello world", "Approved registration", pass: "22/02/2017" Any date Me returns this ;Delimiter must not be alphanumeric or backslash in C: wamp www ecoprintQ ecoLicenseLayout json dados-partners-repots.php on line <i>14
– gabrielfalieri
Failed to put the delimiters in regex, leave so:
preg_match( "#\d{1,2}/\d{1,2}/\d{4}$#"– rray
Basically it’s a syntax error in Regex, I updated the tags.
– Bacco