Question

你好，我有一个php页面，通过ajax请求媒体使用div输出我的html页面中的结果，

我打算使用listviews来推广这个结果，因为我的应用程序针对手机和平板电脑，所以我正在使用jquery移动框架。

我已成功使用此

输出结果

   <?php header('Access-Control-Allow-Origin: *'); ?>
<?php

$users = ($_GET['users']);
$user = ($_GET['user']);



$con = mysqli_connect('localhost','whitech3_manage','famakin','whitech3_manage');
if (!$con) {
  die('Could not connect: ' . mysqli_error($con));
}


mysqli_select_db($con,"whitech3_manage");
$sql="SELECT * FROM data WHERE fromwhere = '".$users."' and towhere = '".$user."'";
$result = mysqli_query($con,$sql);
echo '<head>';
echo "<link rel='stylesheet' href='themes/transport.min.css' />";
echo "<link rel='stylesheet' href='themes/jquery.mobile.icons.min.css' />";
echo "<link rel='stylesheet' href='themes/jquery.mobile.structure-1.4.2.min.css' />";
echo "<script src='themes/jquery-1.10.2.min.js'></script>";
echo "<script src='themes/jquery.mobile-1.4.2.min.js'></script>";

echo '</head>';




while($row = mysqli_fetch_array($result)) {
echo "<div data-role='collapsible' data-theme='b' data-content-theme='b'>";

    echo "<ul data-role='listview' data-filter='true'>";
       echo "<li><a href='#'>". $row['fromwhere'] . "</a></li>";
          echo "<li><a href='#'>". $row['towhere'] . "</a></li>";
             echo "<li><a href='#'>". $row['details'] . "</a></li>";
     echo "<li><a href='#'>". $row['time frame'] . "</a></li>";


 echo "</ul>";
echo "</div>";
}

mysqli_close($con);
?>

但是每当我的结果显示时，他们都没有正确格式化，看起来像列表视图，因此我知道我应该加载我的样式表和javascript标签，以确保列表视图正确显示。

我已经尝试过并使用echo语句来回显我的样式表和js的链接，但它似乎让我的listview正确显示。

这是我试过的

echo '<head>';
echo "<link rel='stylesheet' href='themes/transport.min.css' />";
echo "<link rel='stylesheet' href='themes/jquery.mobile.icons.min.css' />";
echo "<link rel='stylesheet' href='themes/jquery.mobile.structure-1.4.2.min.css' />";
echo "<script src='themes/jquery-1.10.2.min.js'></script>";
echo "<script src='themes/jquery.mobile-1.4.2.min.js'></script>";

echo '</head>';

function showUser() {
    if (showUser == "") {
        document.getElementById("txtHint").innerHTML = "";

        return;
    } else { 
        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) {
                document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
            }
        }
        var usersvalue=encodeURIComponent(document.getElementById("users").value)
var uservalue=encodeURIComponent(document.getElementById("user").value)
        xmlhttp.open("GET","http://whitechapelandpartners.com/getuserwrong.php?users="+usersvalue+"&user="+uservalue, true);


        xmlhttp.send();

    }
}

<div id="txtHint"><b>Person info will be listed here.</b></div><br /><br /><br /><br />
    </div>

任何建议请按照逻辑我希望echo语句能够正常工作

Answer 1

试试这个：

function showUser() {
    if (showUser == "") {
        document.getElementById("txtHint").innerHTML = "";

        return;
    } else { 
        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 collapsible = xmlhttp.responseText;
                document.getElementById("txtHint").innerHTML = $(collapsible).collapsible();

            }
        }
        var usersvalue=encodeURIComponent(document.getElementById("users").value)
var uservalue=encodeURIComponent(document.getElementById("user").value)
        xmlhttp.open("GET","http://whitechapelandpartners.com/getuserwrong.php?users="+usersvalue+"&user="+uservalue, true);


        xmlhttp.send();

    }
}

<div id="txtHint"><b>Person info will be listed here.</b></div><br /><br /><br /><br />
    </div>

我希望这有效。：d

Answer 2

这是我的建议

<?php header('Access-Control-Allow-Origin: *'); ?>
<!DOCTYPE html>
<html>
<head>
<link rel='stylesheet' href='themes/transport.min.css' />
<link rel='stylesheet' href='themes/jquery.mobile.icons.min.css' />
<link rel='stylesheet' href='themes/jquery.mobile.structure-1.4.2.min.css' />
<script type="text/javascript" src='themes/jquery-1.10.2.min.js'></script>
<script type="text/javascript" src='themes/jquery.mobile-1.4.2.min.js'></script>
<script type="text/javascript">

function showUser(user) { // how do you call this? You need to pass it a user too
    if (user == "") {
        $("#txtHint").empty();
        return;
    } 
    var usersvalue=encodeURIComponent($"#users").val()),
        uservalue=encodeURIComponent($("#user").val());
    // assuming you are accessing the same origin
    $.get("/getuserwrong.php?users="+usersvalue+"&user="+uservalue,
      function(data) {
        $("#txtHint").html(data);
      }
    ); 
}

</script>
</head>
<body>
<?php
$users = ($_GET['users']);
$user = ($_GET['user']);
// I think you do not want to show your userID and password here on this site
$con = mysqli_connect('localhost','...','...','...'); 
if (!$con) {
  die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"whitech3_manage");
$sql="SELECT * FROM data WHERE fromwhere = '".$users."' and towhere = '".$user."'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result)) {  
  echo <<<EOT 
    <div data-role='collapsible' data-theme='b' data-content-theme='b'>
      <ul data-role='listview' data-filter='true'>
        <li><a href='#'>$row['fromwhere']</a></li>
        <li><a href='#'>$row['towhere']</a></li>
        <li><a href='#'>$row['details']</a></li>
        <li><a href='#'>$row['time frame']</a></li>
      </ul>
   </div>
EOT;   
}
mysqli_close($con);
?>

<div id="txtHint"><b>Person info will be listed here.</b></div>
</body>
</html>

如何使用php回显样式和脚本标记

2 个答案: