PHP remove portion of a string by pattern

Question

i have bunch of strings, where i need cut out one portion:

1. Adaptation of a one-step worst-case optimal univariate algorithm of bi-objective Lipschitz optimization to multidimensional problems, vol. 21 [looked: 2014 m. september 15 d.], p. 89-98.
2. On benchmarking stochastic global optimization algorithms, 2014 [looked 2014 m. spetember 17 d.], p. 9-12.
3. Quadratic programming with complementarity constraints for multidimensional scaling with city-block distances, Vol. 2, no. 4 [looked 2014 m. december 05 d.], p. 248-259.

and many others. I need remove [looked YYYY m. month dd d.] portion. Which begins with "[looked" and ends with "]". How can i achieve this?

i have tried preg_replace, but i cannot build up great working pattern.

Avinash Raj · Accepted Answer · 2015-10-29 07:31:53Z

1

You may try this,

preg_replace('~\[Looked\b[^\]]*\]~', '', $str);

answered Oct 29, 2015 at 7:31

Avinash Raj

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Avinash Raj Over a year ago

add \s* before the first \[ , if you want to remove also the leading spaces.

Andrius · Accepted Answer · 2015-10-29 07:34:13Z

1

This works:

preg_replace("/\[looked(.+?)\]/","",$str)

answered Oct 29, 2015 at 7:34

Andrius

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