Question

我创建了proba.html文件，getuser.php，名为'mobilni'的数据库，以及名为'imena'的表。代码出了什么问题，它没有显示表中的结果，只有空白？

proba.html代码：

<script>
function showUser(str) {
    if (str == "") {
        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;
            }
        }
        xmlhttp.open("GET","getuser.php?q="+str,true);
        xmlhttp.send();
    }
}
</script>

<form>
<select name="okrug" onchange="showUser(this.value)">
  <option value="">Okrug:</option>
  <option value="Raski">Raski</option>
  <option value="Banatski">Banatski</option>
  <option value="Backi">Backi</option>
  <option value="Beogradski">Beogradski</option>
  </select>
</form>
<br>
<div id="txtHint"><b>Person info will be listed here...</b></div>

getuser.php代码：

<?php
$q = intval($_GET['q']);

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

mysqli_select_db($con,"mobilni");
$sql="SELECT * FROM imena WHERE Okrug = '".$q."'";
$result = mysqli_query($con,$sql);

echo "<table>
<tr>
<th>Ime</th>
<th>Okrug</th>
<th>Telefon</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
    echo "<tr>";
    echo "<td>" . $row['Ime'] . "</td>";
    echo "<td>" . $row['Okrug'] . "</td>";
    echo "<td>" . $row['Telefon'] . "</td>";
    echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>

Answer 1

而不是

$q = intval($_GET['q']);

试试这个：

$q = $_GET['q'];

看，如果有帮助。

Php和Ajax没有显示SQL的结果

1 个答案: