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[SOLVED] –             if(user.charAt(x + 1) == ‘y’)

by admin | Aug 29, 2022 |             if(user.charAt(x + 1) == 'y')

[SOLVED] –             if(user.charAt(x + 1) == ‘y’)

by admin | Aug 13, 2022 |             if(user.charAt(x + 1) == 'y')

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