1

我无法在我的 php / mysql Select 查询中获取全局变量。

这是我目前拥有的代码，它不会从我的数据库中返回任何值。

<HTML>
<HEAD>
<SCRIPT type="text/javascript">
window.onload = function()
{
    document.getElementById('wall').innerHTML="<?php zoomLevelOne();?>";
}
</SCRIPT>
<?php
$startBrick = rand (1, 4);
function zoomLevelOne()
{
    global $startBrick;
    $brickNo = rand (1, 4);
    $con = mysql_connect("localhost","root","password");
    if(!$con){die('Could not connect: ' . mysql_error());}
    mysql_select_db("wall", $con);
    $result = mysql_query("SELECT * FROM bricks WHERE BrickNo=$startBrick");
    $brick = mysql_fetch_array($result);
    mysql_close($con);
    echo $brick['year'];
}
?>
</HEAD>
<BODY>
<DIV id='wall'>
</DIV>
</BODY>
</HTML>

如果我将选择查询的结尾更改为：

WHERE BrickNo=$brickNo");

所以整个代码是这样的：

<HTML>
<HEAD>
<SCRIPT type="text/javascript">
window.onload = function()
{
    document.getElementById('wall').innerHTML="<?php zoomLevelOne();?>";
}
</SCRIPT>
<?php
$startBrick = rand (1, 4);
function zoomLevelOne()
{
    global $startBrick;
    $brickNo = rand (1, 4);
    $con = mysql_connect("localhost","root","password");
    if(!$con){die('Could not connect: ' . mysql_error());}
    mysql_select_db("wall", $con);
    $result = mysql_query("SELECT * FROM bricks WHERE BrickNo=$brickNo");
    $brick = mysql_fetch_array($result);
    mysql_close($con);
    echo $brick['year'];
}
?>
</HEAD>
<BODY>
<DIV id='wall'>
</DIV>
</BODY>
</HTML>

这对我来说似乎是正确的，因为我所做的只是将局部变量更改为全局变量，但它不起作用，所以也许我做错了什么，或者只是我的编码不佳。

谁能帮我。

4

1 回答 1

0

我的答案看起来很简单。代码在声明变量之前调用函数。只需将脚本块移动到定义函数之后，就在标记之前。

<HTML>
<HEAD>
<?php
$startBrick = rand (1, 4);
function zoomLevelOne()
{
    global $startBrick;
    $brickNo = rand (1, 4);
    $con = mysql_connect("localhost","root","password");
    if(!$con){die('Could not connect: ' . mysql_error());}
    mysql_select_db("wall", $con);
    $result = mysql_query("SELECT * FROM bricks WHERE BrickNo=$startBrick");
    $brick = mysql_fetch_array($result);
    mysql_close($con);
    echo $brick['year'];
}
?>
<SCRIPT type="text/javascript">
window.onload = function()
{
    document.getElementById('wall').innerHTML="<?php zoomLevelOne();?>";
}
</SCRIPT>
</HEAD>
<BODY>
<DIV id='wall'>
</DIV>
</BODY>
</HTML>
于 2014-03-06T15:18:11.877 回答