Question

我想知道是否有人可以帮助我。

我找到this教程，使用mySQL表数据创建一个下拉菜单，然后返回页面上的结果列表。

按照本教程，我将数据库中的表和所需的脚本放在一起，如下所示：

搜索表单

<html>
<head>

<script type="text/javascript">
function ajaxFunction()
{

//document.writeln(val)
var httpxml;
try
{
// Firefox, Opera 8.0+, Safari
httpxml=new XMLHttpRequest();
}
catch (e)
{
// Internet Explorer
try
{
httpxml=new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e)
{
try
{
httpxml=new ActiveXObject("Microsoft.XMLHTTP");
}
catch (e)
{
alert("Your browser does not support AJAX!");
return false;
}
}
}

function stateChanged(){
if(httpxml.readyState==4){

var myObject = eval('(' + httpxml.responseText + ')');


//var myObject = httpxml.responseText;
//document.getElementById("display").innerHTML=myObject;





var msg=myObject.value[0].message;
if(msg.length > 0){document.getElementById("msg").innerHTML=msg;}
else{document.getElementById("msg").style.display='none';}

var str="<table width='50%' bgcolor='#ffffff' align=center><tr><th>ID</th><th>Name</th></tr>";
var color="#f1f1f1";
for(i=0;i<myObject.data.length;i++)
{ 
if((i%2)==0){color="#ffffff";}
else{color="#f1f1f1";}
str = str + "<tr bgcolor="+color+"><td>" + myObject.data[i].subcat_id + " </td><td>"+ myObject.data[i].subcat_name + "</a></td></tr>"

}



str = str + "</table>" ;


document.getElementById("display").innerHTML=str;
}
}

var url="subcat2.php";
var cat_id=document.myForm.cat_id.value;
url=url+"?cat_id="+cat_id;

url=url+"&kid="+Math.random();
//alert(url)
httpxml.onreadystatechange=stateChanged;
httpxml.open("GET",url,true);
httpxml.send(null);
// document.getElementById("display").innerHTML="Please Wait....";
document.getElementById("msg").style.background='#f1f1f1';
document.getElementById("msg").innerHTML="Please Wait ... ";
document.getElementById("msg").style.display='inline';
}

</script>

<?php
require "z_db.php";
echo "</head><body onload="ajaxFunction()";>";
echo "<center><table border='0' width='100%' cellspacing='0' cellpadding='0' > <tr bgcolor='#ffffcc'><form name=myForm method='post' onSubmit="ajaxFunction(this.form); return false">


<td align='center' colspan=2><font face='verdana, arial, helvetica' size='2' ><b> Select a Category</b> </font></td></tr>";
echo "<tr>";
echo "<td align='center'>";
$query="SELECT * FROM plus2_cat order by cat_name";
$result=mysql_query($query);
echo mysql_error();
echo "<select name=cat_id onChange="ajaxFunction()"><option value=0>Show All</option>";
while($nt=mysql_fetch_array($result)){
echo "<option value=$nt[cat_id]>$nt[cat_name]</option>";
}

echo "</select>";
echo "</font></td>";
echo "</tr></form>";
echo "</table>";


?>
<div id=msg style="position:absolute; z-index:1; left: 1100px; top: 0px;" >This is message area</div>

<div id="display"><b>Records will be displayed here</b></div>

</body>
</html>

subcat2.php

<?php
////// Your Database Details here /////////
require ("z_db.php"); 
//Your database details here
//////////////////////////// Main Code sarts ///////////////
$cat_id=$_GET['cat_id'];
//$cat_id=11;

if(!is_numeric($cat_id)){
$message.="Data Error |";
exit;
}

//$message.=" :Status = $status";
if($cat_id>0){
$q=mysql_query("select subcat_id, subcat_name from plus2_subcat where cat_id=$cat_id order by subcat_name");
}else{
$q=mysql_query("select subcat_id, subcat_name from plus2_subcat order by subcat_name ");
$cat_id=0;
}
//$message .=" Cat_id=$cat_id ";
$message .= mysql_error();

$str= "{ "data" : [ ";

while($nt=mysql_fetch_array($q)){
$str=$str."{"subcat_id" : "$nt[subcat_id]", "subcat_name" : "$nt[subcat_name]"},";
//$str=$str."{"myclass" : "$nt[class]"},";

}
$str=substr($str,0,(strLen($str)-1));
$message=$message. " Records displayed";
$str=$str."],"value" : [{"cat_id" : $cat_id,"message" : "$message"}]}";
//echo json_encode($str);

echo $str;
?>

z_db.php

<?php
mysql_connect("host", "user", "password")or
die(mysql_error());
mysql_select_db("database");
?>

我的问题是，当我尝试运行它时，我收到以下错误： Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /homepages/2/d333603417/htdocs/development/catsearch.php on line 91这是搜索表单中的这一行：echo "</head><body onload="ajaxFunction()";>";。

我必须承认我已经猜到'z_db.php'文件的结构应该是这样的，因为这没有显示所以这可能是问题所在。

我只是想知道是否有人可以看看这个请让我知道我哪里出错了。

Answer 1

尝试：

echo "</head><body onload='ajaxFunction()';>";

您不能嵌套单/双引号。

Ajax菜单检索mySQL记录

1 个答案: