Php和Ajax未顯示SQL的結果

Question

我創建了proba.html文件getuser.php，數據庫名爲'mobilni'，表名爲'imena'。代碼有什麼問題，它沒有顯示錶的結果，只有空白？

proba.html代碼：

<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代碼：

<?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']); 

試試這個：

$q = $_GET['q']; 

看，如果有幫助。