在表格外显示信息的AJAX下拉选择站点

Question

下拉选择将填充正确的数据，但输出始终显示在表外部。 我的代码有什么问题？ 有什么想法/建议吗？ 提前致谢，

在此处输入图片说明

select.php

<?php
$q = $_GET['q'];

$con = mysqli_connect('host','username','password','database');
if (!$con)
{
    die('Could not connect: ' . mysqli_error($con));
}

mysqli_select_db($con,"ajax_demo");

$sql="SELECT * FROM timeSheet, timeSheetUsers WHERE timeSheet.userName=timeSheetUsers.userName AND timeSheet.userName= '".$q."' ORDER BY startTime ASC";

$result = mysqli_query($con,$sql);

echo "<table border ='1'>
<tr>
<th bgcolor='#F0E68C'>username</th>
<th bgcolor='#E0FFFF'>date</th>           
<th bgcolor='#F0E68C'>starTime</th>
<th bgcolor='#E0FFFF'>endTime</th>
<th bgcolor='#F0E68C'>total</th>   
</tr>";

    echo "<br />\n";
    echo "<br />\n";
    echo "<br />\n";
    echo "<br />";
    echo "<br />";
    echo "<br />";
                                                            //  

while($row = mysqli_fetch_array($result))
{
    echo "<tr>";
    echo "<tdbgcolor='#F0E68C'>" . $row['userName'] . "</td>";
    echo "<tdbgcolor='#F0E68C'>" . $row['date'] . "</td>";
    echo "<tdbgcolor='#F0E68C'>" . $row['startTime'] . "</td>";
    echo "<tdbgcolor='#F0E68C'>" . $row['endTime'] . "</td>";
    echo "<tdbgcolor='#F0E68C'>" . $row['total'] . "</td>";

    echo "</tr>";
}
echo "</table>";

mysqli_close($con);
?>

home.php

<html>
<head>

<script type="text/javascript"
    src="http://code.jquery.com/jquery-1.11.0.min.js"></script>

<script>
function showUser(str)
{
if (str=="")
  {
  document.getElementById("txtHint").innerHTML="";
  return;
  } 
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();
  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
     var txtHint = document.getElementById("txtHint");
     $("#txtHint").css("display", "none");
     $("#txtHint").fadeIn('slow');
    txtHint.innerHTML=xmlhttp.responseText;
    }
  }
xmlhttp.open("GET","select.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>

<?php
require "credentials.php";
try {
    $database = "database"; 
    $conn     = new PDO("mysql:host=$hostname;dbname=$database", $username, $password);
    echo "<p>Database Connection OK.<p>";
}
catch (PDOException $e) {
    $conn->setArribute(PDO::ATT_ERRMODE, PDO::ERRMODE_EXCEPTION);
    echo "Could not open database.";
}

echo "<form>";
echo "<select name='selStudentWorker' onChange='showUser(this.value)'>";
$sql = $conn->prepare("SELECT firstName, lastName, userName FROM timeSheetUsers ORDER BY firstName ASC");
if ($sql->execute(array(
    $_SESSION['userName']
))) {
    while ($row = $sql->fetch()) {
        echo "<option value='".$row['userName']."' "."selected='selected'>" . $row['firstName'] . " " . $row['lastName'] . " (" . $row['userName'] . ")";
        echo "</option>";

    }
}
echo "</select>";
echo "</form>";
?>
</body>
</html>

参考：基于此的代码-> http://www.w3schools.com/ajax/ajax_database.asp

Answer 1

您的td标签中有一个错误：

<tdbgcolor='#F0E68C'>

应该：

<td bgcolor='#F0E68C'>

在表格外显示信息的AJAX下拉选择站点

问题描述

1 个解决方案

解决方案1
0 已采纳 2014-11-25 19:49:12

在表格外显示信息的AJAX下拉选择站点

问题描述

1 个解决方案

解决方案1 0 已采纳 2014-11-25 19:49:12

解决方案1
0 已采纳 2014-11-25 19:49:12