php里面的unset 可以把值置空，但是并不会改变值已经有的索引位置
1 $testArr=array('frist','second','third'); 2 echo '<pre>'; 3 echo '<br/>---------------------------------proto ------------------------------</br>'; 4 print_r($testArr); 5 foreach($testArr as $key=>$val){ 6 unset($testArr[$key]); 7 } 8 echo '<br/>---------------------opo---------------------------</br>'; 9 print_r($testArr); 10 $testArr[]='good!'; 11 echo '<br/>---------------------add---------------------------</br>'; 12 print_r($testArr); 13 echo '</pre>';
If a static variable is unset() inside of a function, unset() destroys the variable only in the context of the rest of a function. Following calls will restore the previous value of a variable.
1 function foo() 2 { 3 static $bar; 4 $bar++; 5 echo "Before unset: $bar, "; 6 unset($bar); 7 $bar = 23; 8 echo "after unset: $bar<br/>"; 9 } 10 11 foo(); 12 foo(); 13 foo(); 14
vld分析： 15 function foos(&$bar) 16 { 17 unset($bar); 18 $bar = "blah"; 19 } 20 21 $bar = 'something'; 22 echo "$bar<br/>"; 23 24 foos($bar); 25 echo "$bar<br/>";
If a variable that is PASSED BY REFERENCE is unset() inside of a function, only the local variable is destroyed. The variable in the calling environment will retain the same value as before unset() was called.
这是手册中发现的，突然感觉把手册再翻一遍，是个正确的选择