php中的变量变量使用WAMP服务器问题

问题描述：

I am running newest version of WAMPServer on Windows 7 Home Premium. It accesses database (remote) fine, it runs basic PHP fine. My issue is with variable variables. On my godaddy server, the exact code runs fine.

When I try to include a file that uses variable variables (in this case $state{$i}) in WAMPServer says:

Notice: Undefined variable: state in E:\wamp\www\CCRN411\inc\state_drop_down.inc.php on line 2.

Note that it is looking for state, not state1 as the code generates for $i= 1 which I declare on the first line of the same file.

<select class="dropdownState" name="<?php $i=1; echo 'state' . $i; ?>" id="<?php echo 'state' . $i; ?>">
<option value="DL" <?php if ($state{$i} == "DL") {echo ('selected="selected"');} ?>>DELETE</option>
<option value="ZZ" <?php if ($state{$i} == "ZZ") {echo ('selected="selected"');} ?>>NONE</option>

If I add

$state1 = . and it is empty. What is the issue with my variable variable?

Help please.

我在Windows 7 Home Premium上运行最新版本的WAMPServer。它访问数据库（远程）很好，它运行基本PHP很好。我的问题是变量变量。在我的godaddy服务器上，确切的代码运行正常。 p>

当我尝试在WAMPServer中包含一个使用变量变量的文件（在本例中为$ state {$ i}）时说： p>

注意：未定义的变量：第2行的E：\ wamp \ www \ CCRN411 \ inc \ state_drop_down.inc.php中的状态。 p>

注意它正在寻找状态，而不是state1，因为代码为$ i = 1生成，我在同一文件的第一行声明。 p>

 ＆lt; select class =“dropdownState”  name =“＆lt;？php $ i = 1; echo'state'。$ i;？＆gt;”  id =“＆lt;？php echo'state'。$ i;？＆gt;”＆gt; 
＆lt; option value =“DL”＆lt;？php if（$ state {$ i} ==“DL”）{echo  （'selected =“selected”'）;}？＆gt;＆gt; DELETE＆lt; / option＆gt; 
＆lt; option value =“ZZ”＆lt;？php if（$ state {$ i} ==“ZZ”）{echo  （'selected =“selected”'）;}？＆gt;＆gt; NONE＆lt; / option＆gt; 
  code>  pre> 
 
 如果我添加
 
  p>  $ state1 =。 它是空的。
我的变量变量有什么问题？ p> 
 
 
请帮助。 p> 
  div>

答

Wrong syntax. You want ${"state$i"}, as in

if (${"state$i"} == "DL") ...

php中的变量变量使用WAMP服务器问题

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