How to replace string while creating variable using GREP search pattern

Question

// constraints are:// XX T will always be first and in caps ==> ^XX T// T will always have a value of max of 2 digits ==> {1,2}// L may or may not be present but if present will always have a value of max of 2 digits ==> {1,2}// the digits of L if present will always be the end of the string ==> (/\d+$/) var testName1 = "XX T3 L22";  // viable upper case search(/^XX T\d{1,2}( L\d{1,2}$)/g)var testName2 = "XX T44 L1"; var testName3 = "XX T5";  // viable upper case search(/^XX T\d{1,2}$/g)var testName4 = "XX T66;  // for testName1 ==> lNumber1 = 22if (testName1.toString().search(/^XX T\d{1,2}( L\d{1,2}$)/g) === 0) lNumber1 = testName1.toString().replace((/^XX T\d{1,2}( L\d{1,2}$)/g), testName1.toString().match(/\d+$/)); // how can I return the T value in all 4 examples?// results:// lNumber1 = 3// lNumber2 = 44// lNumber3 = 5// lNumber4 = 66

m1b · Accepted Answer

Hi @nutradial, I assume you are doing this via ExtendScript... see if this helps. The match method of string is great for this.

- Mark

var tests = [
    "XX T3 L22",
    "XX T44 L1",
    "XX T5",
    "XX T66"
];

// match two digits of T and two digits of L (if L exists)
var matcher = /^XX T(\d{1,2})(?: L(\d{1,2}))?/;

for (var i = 0, T, L; i < tests.length; i++) {

    // perform the regex match
    var match = tests[i].match(matcher);

    if (!match)
        // no match at all
        continue;

    T = match.length > 0 ? Number(match[1]) : undefined;
    L = match.length > 1 ? Number(match[2]) : undefined;

    $.writeln('test string "' + tests[i] + '":  T = ' + T + '  L = ' + L);

}

nutradial · Answer

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