Tec*_*ude 1 php variables scope switch-statement
在PHP中,我有一个名为$ chickens的开关之外的值,等于1.在通过switch语句之后,我希望将值更改为b.
//original value of var
$chickens = 1;
switch ($chickens) {
case "0":
$chickens === "a";
break;
case "1":
$chickens === "b";
break;
case "2":
$chickens === "c";
break;
case "3":
$chickens === "b";
break;
}
//want this to be "b" defined in switch
echo $chickens;
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由于某种原因,价值不会更新:(.任何想法
你使用的是错误的操作员.=是赋值运算符.===是一个比较运算符:
//original value of var
$chickens = 1;
switch ($chickens) {
case "0":
$chickens = "a";
break;
case "1":
$chickens = "b";
break;
case "2":
$chickens = "c";
break;
case "3":
$chickens = "d";
break;
}
//want this to be "b" defined in switch
echo $chickens;
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