问题描述
请问我的代码有问题....它不会从数据库返回数据。
please am having problem with my code.... it's not returning data from database.
<script type="text/javascript">
var myTimer;
function ajax_json_data(){
var databox = document.getElementById("databox");
var arbitrarybox = document.getElementById("arbitrarybox");
var hr = new XMLHttpRequest();
hr.open("POST", "request.php", true);
hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
hr.onreadystatechange = function() {
if(hr.readyState == 4 && hr.status == 200) {
var d = JSON.parse(hr.responseText);
arbitrarybox.innerHTML = d.arbitrary.returntime;
databox.innerHTML = "";
for(var o in d){
if(d[o].title){
databox.innerHTML += '<p><a href="page.php?id='+d[o].id+'">'+d[o].title+'</a><br>';
databox.innerHTML += ''+d[o].cd+'</p>';
}
}
}
}
hr.send("limit=");
databox.innerHTML = "requesting...";
myTimer = setTimeout('ajax_json_data()',6000);
}
</script>
<?php
header("Content-Type: application/json");
if(isset($_POST['limit'])){
$limit = preg_replace('#[^0-9]#', '', $_POST['limit']);
include_once("phpincludes/dataconect.php");
$i = 0;
$jsonData = '{';
$sqlString = "SELECT * FROM bible WHERE ORDER BY RAND() LIMIT $limit";
$query = mysqli_query($db_conx, $sqlString) or die (mysqli_error());
while ($row = mysqli_fetch_array($query, MYSQLI_ASSOC)) {
$i++;
$id = $row["id"];
$title = $row["title"];
$cd = $row["creationdate"];
$cd = strftime("%B %d, %Y", strtotime($cd));
$jsonData .= '"article'.$i.'":{ "id":"'.$id.'","title":"'.$title.'", "cd":"'.$cd.'" },';
}
$now = getdate();
$timestamp = $now[0];
$jsonData .= '"arbitrary":{"itemcount":'.$i.', "returntime":"'.$timestamp.'"}';
$jsonData .= '}';
echo $jsonData;
}else
echo ("nothing was posted");
?>
推荐答案
_POST [' limit']））{ _POST['limit'])){
limit = preg_replace（' ＃[^ 0-9 ]＃'，' '， limit = preg_replace('#[^0-9]#', '',
_POST [' limit']）; include_once（ phpincludes / dataconect.php）; _POST['limit']); include_once("phpincludes/dataconect.php");