我对 javascript/jquery 很陌生，一直在研究 w3schools 教程和 jquery 文档，并制作了一个页面，该页面将接受一些用户输入，然后通过 javascript 打印一些输出。我以为我可以修改它以按顺序工作，但是当第二页应该调用第三页时我得到
ReferenceError: $gp is not defined
[Break On This Error]
$('#txtField2').load('getglycosites.php?q='+$gp+'r='+$('.glycotype').val());
如果有人有能够在 jquery 脚本中使用 $_get 变量的技巧，我将不胜感激。
页面代码:
首页(test.php):
<html>
<head>
<title>LeidenGlycoPeptide DataBase</title>
<script src="jquery-1.9.1.min.js"></script>
</head>
<script>
$(document).ready(function() {
$('.glycoprotein').change(function() {
$('#txtField').load('getglycopeptides.php?gp='+$('.glycoprotein').val());
});
});
</script>
<body>
<h1>Welcome to the LeidenGlycoPeptide DataBase</h1>
<?php
$link = mysql_connect("localhost","reader","") or die (mysql_error());
mysql_select_db('leidenGlycoPeptide') or die ();
$query = 'select protein from glycoPeptide';
$result = mysql_query($query);
mysql_close($link);
?>
<form>
<p>Select glycopeptide to search for (interactive dialog)</p>
<?php
echo"<select class=\"glycoprotein\">";
echo"<option value=\"\">select glycoprotein</option>";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
foreach ($line as $col_value) {
echo"<option value=\"$col_value\">$col_value</option>";
}
}
echo"</select>";
?>
</form>
<br>
<div id="txtField"></div>
</body>
</html>
第二页(get糖肽.php):
<html>
<head>
<title>glyco</title>
<script src="jquery-1.9.1.min.js"></script>
</head>
<script>
$(document).ready(function() {
$('.glycotype').change(function() {
//NOTE $q/q are undefined
$('#txtField2').load('getglycosites.php?q='+$gp+'r='+$('.glycotype').val());
});
});
</script>
<body>
<?php
// The next variable is the one I would like to use inside the jquery
$gp=$_GET["gp"];
$link = mysql_connect("localhost","reader","") or die (mysql_error());
mysql_select_db("leidenGlycoPeptide",$link) or die();
$query = "select glycoType from run,glycoPeptide where run.id = glycoPeptide.id and glycoPeptide.protein like '".$gp."'";
$result = mysql_query($query);
?>
<form>
<?php
echo "<select class=\"glycotype\">";
echo "<option value=\"\">select glycosylation</option>";
while ($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
foreach ($row as $col_value)
{
echo"<option value=\"$col_value\">$col_value</option>";
}
}
echo "</select>";
mysql_close($link);
?>
</form>
<br>
<div id="txtField2"></div>
</body>
</html>
第三页(getglusites.php):
<html>
<head>
<title>sites</title>
</head>
<body>
<?php
$gp=$_GET["gp"];
$r=$_GET["r"];
$link = mysql_connect("localhost","reader","") or die (mysql_error());
mysql_select_db("leidenGlycoPeptide",$link) or die();
$query = "select glycoSite from run,glycoPeptide where run.id = glycoPeptide.id and run.glycoType like '".$r."' and glycoPeptide.protein like '".$q."'";
//echo $query;
$result = mysql_query($query);
echo "<select name=\"site\" onchange=\"foo\">";
echo "<option value=\"\">select glycosite</option>";
while ($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
foreach ($row as $col_value)
{
echo"<option value=\"$col_value\">$col_value</option>";
}
}
echo "</select>";
mysql_close($link);
?>
</body>
</html>
PS:第三页目前只应该显示一个新的选择字段，但一旦我开始工作，它将成为一个大顺序序列的一部分。
提前致谢
最佳答案
我能够通过添加以下行(到 jquery 脚本)来“修复”此问题:
var gp = '<?php echo htmlspecialchars($_GET['gp']); ?>';
使总脚本成为第三页的一部分:
<script>
var gp = '<?php echo htmlspecialchars($_GET['gp']); ?>';
$(document).ready(function() {
$('.glycotype').change(function() {
$('#txtField2').load('getglycosites.php?q='+gp+'&r='+$('.glycotype').val());
});
});
</script>
关于javascript - 顺序jquery 'pages'，第三帧给出引用错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/14850138/