note 63920 deleted from language.operators.logical by nlopess

From: Date: Fri, 02 Jun 2006 08:55:52 +0000
Subject: note 63920 deleted from language.operators.logical by nlopess
References: 1  Groups: php.notes 
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Note Submitter: jhibbard at gmail dot com ---- I have read a few of these responses and quite honestly didn't find one that explained the differences between the "||" and "OR" operators. The best way to explain it is with an example. <?php echo "TEST 1:"; $choice1 = false; $choice2 = "dog"; $default = "other"; $val = ($choice1 != false ? 'hah' : 'boo') || ($choice2 != false ? 'hah2' : 'boo2') || ($default != false ? 'hah3' : 'boo3'); echo $val; echo "<BR /><BR /><BR />TEST 2:"; $choice1 = false; $choice2 = "dog"; $default = "other"; //$val = $choice1 or $choice2 or $default; $val = ($choice1 != false ? 'hah' : 'boo1' or ($choice2 != false ? 'hah2' : 'boo2') or ($default != false ? 'hah3' : 'boo3'); //$val = $choice1 or $choice2 or $default; echo $val; ?> If you use this code, you will have 2 responses. The first will be TEST 1:1 and the second will be TEST 2:boo1 The reason is because how the 2 operators respond. The "||" statement returns boolean (true or false), thus the return value of 1. The "OR" statement on the other hand, returns the actual statement on TRUE, thus why we are getting boo1. Now, let's give one more example and be done with it. If we use these statements : <?php //Statement ONE with || $sql = "SELECT * FROM table"; $rst = myqsl_query($sql) || die(MySQL_Error()); //Statement TWO with OR $sql = "SELECT * FROM table"; $rst = myqsl_query($sql) OR die(MySQL_Error()); ?> You will notice what I mean. Statement one will return true (or simply 1). Statement 2 will return the actual result of the query (usually Resource ID #5 or the actual mysql error). I hope this makes it more clear to the newbie/apprentice who is looking for a deeper explanation of these logical operators.

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