我有这个程序，我一直在努力。当我使用此代码加载页面时，它完美无缺。我添加了一个即时搜索输入，每次我尝试即时搜索它失败。它为我提供了unexpected end of input input，Unexpected token S或Unexpected token <
这是完整的代码，注释掉了返回数据，我试图找出我出错的地方。
html Inputs:
<form>
<div class="left group">Search Title: <input type="text" name="title" id="title" class="css-input" onkeyup="searchdb(this.id);" />|</div>
<div class="left group">Search Stars: <input type="text" name="stars" id="stars" class="css-input" onkeyup="searchdb(this.id);" />|</div>
<div class="left group">Search View Key: <input type="text" name="vks" id="vks" class="css-input"onkeyup="searchdb(this.id);" />|</div>
<br/><br/>
<div class="line1"></div>
<br/>
</form>
完整的Javascript：
var ttl='ttl';
var str='str';
function listedLoad(ttl,str){
var vid_gen = '';
var x = ttl;
var y = str;
var data = 'pass='+x+'&search='+y;
var dir = 'add/video_php/';
var video_select = _("list");
var create = new XMLHttpRequest();
create.open("POST", "php/list_view.php", true);
create.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
create.onreadystatechange = function() {
if(create.readyState == 4 && create.status == 200) {
//console.log(create.responseText); //
//_("list").innerHTML = create.responseText;// --> All data returns correct.
//return false; //
var d = JSON.parse(create.responseText);
for(var o in d){
var div = document.createElement('div');
div.className = 'item-group';
var item1 = document.createElement('div');
item1.className = 'item-id left';
item1.innerHTML = 'Id: '+d[o].id;
var item2 = document.createElement('div');
item2.className = 'item-uid left';
item2.innerHTML = 'Uid: '+d[o].uid;
var item3 = document.createElement('div');
item3.className = 'item-title';
item3.innerHTML = 'Title: '+d[o].title;
var item4 = document.createElement('div');
item4.className = 'item-dur right';
item4.innerHTML = 'Dur: '+d[o].dur;
var item5 = document.createElement('div');
item5.className = 'item-year';
item5.innerHTML = 'Year: '+d[o].year;
var item6 = document.createElement('div');
item6.className = 'item-stars';
item6.innerHTML = 'Stars: '+d[o].stars;
var item7 = document.createElement('div');
item7.className = 'item-genere';
item7.innerHTML = 'Genere: '+d[o].genere;
var item8 = document.createElement('div');
item8.className = 'item-desc';
item8.innerHTML = 'Desc: '+d[o].desc;
var item9 = document.createElement('div');
item9.className = 'item-loc';
item9.innerHTML = 'Loc: '+d[o].loc;
var item10 = document.createElement('div');
item10.className = 'item-udate';
item10.innerHTML = 'Udate: '+d[o].udate;
var item11 = document.createElement('div');
item11.className = 'item-vdate';
item11.innerHTML = 'Vdate: '+d[o].vdate;
var item12 = document.createElement('div');
item12.className = 'item-vcount left';
item12.innerHTML = 'Vcount: '+d[o].vcount;
var item13 = document.createElement('div');
item13.className = 'item-vrate left';
item13.innerHTML = 'Vrate: '+d[o].vrate;
_("list").appendChild(div);
div.appendChild(item9);
div.appendChild(item1);
div.appendChild(item2);
div.appendChild(item12);
div.appendChild(item13);
div.appendChild(item3);
item3.appendChild(item4);
item3.appendChild(item5);
div.appendChild(item6);
div.appendChild(item7);
div.appendChild(item8);
div.appendChild(item10);
div.appendChild(item11);
}
}
}
create.send(data);
}
function searchdb(x){
if(x == 'title'){title = _('title').value;ttl='title';str=title;}
else if(x == 'stars'){stars = _('stars').value;ttl='stars';str=stars;}
else if(x == 'vks'){vks = _('vks').value;ttl='vks';str=vks;}
console.log(ttl+' '+str); // --> all variables return correctly.
listedLoad(ttl,str);
}
PHP：
<?php
header("Content-Type: application/json");
error_reporting(E_ALL);
include_once '../../admin/admin-temp-php/db_conn.php';
include_once '../../add/video_php/vid_functions.php';
//echo $_POST['pass'];echo $_POST['search'];exit(); --> Here i echoed data back to the page. Comes in correct.
if($_POST['pass'] == 'undefined'){ //
$_POST['search']=''; //
$searchLoc = '*'; //
$srh = 'videos'; // --> On page load, this works perfect. No issues here
$where = ''; //
$search = 'title'; //
} //
//////////////////////////////////////////////////
if($_POST['pass'] == 'title'){ //
$searchLoc = '*'; //
$srh = 'videos'; //
$where = ''; //
$search = 'title'; //
} //
if($_POST['pass'] == 'stars'){ //
$find=$_POST["search"]; //
$searchLoc = '*'; //
$srh = 'videos'; //
$where = " WHERE stars LIKE '\%$find\%' "; // -->this is where i think my problem lays
$search = 'stars'; //
} //
if($_POST['pass'] == 'vks'){ //
$find=$_POST["search"]; //
$searchLoc = '*'; //
$srh = 'videos'; //
$where = " WHERE loc LIKE '\%$find\%' "; //
$search = 'loc'; //
} //
//////////////////////////////////////////////////
$sql = "SELECT ".$searchLoc." FROM ".$srh.$where." ORDER BY ".$search." ASC";
//echo $sql;exit();// <-- the $sql syntax looks good here in all inputs
$query = mysqli_query($db_conn, $sql) or die (mysqli_error());
$data = '{';
$i = 0;
if($query){
while ($row = mysqli_fetch_array($query)) {
$id = $row['id'];
$uid = $row['uid'];
$title = $row['title'];
$dur = $row['dur'];
$year =$row['year'];
$stars = $row['stars'];
$genere = $row['genere'];
$description = $row['description'];
$loc = $row['loc'];
$cover = $row['cover'];
$udate = $row['udate'];
$v_date = $row['v_date'];
$v_count = $row['v_count'];
$v_rate = $row['v_rate'];
$ip = $row['ip'];
$explode = explode('/',$loc);
$loc = explode('.',$explode[2]);
$loc = $loc[0];
$data .= '"mov'.$i.'":{ "id":"'.$id.'","uid":"'.$uid.'", "title":"'.$title.'", "dur":"'.$dur.'", "year":"'.$year.'", "stars":"'.$stars.'", "genere":"'.$genere.'", "desc":"'.$description.'", "loc":"'.$loc.'", "udate":"'.$udate.'", "vdate":"'.$v_date.'", "vcount":"'.$v_count.'", "vrate":"'.$v_rate.'" },';
$i++;
}
$data = chop($data,',');
$data .= '}';
echo $data;
}else{
echo 'DATA FAILED TO LOAD';
exit();
}
mysqli_free_result($query);
?>
我一直在尝试许多不同的方式。我相信我的问题在PHP if语句的开头$sql。第一个块完美无缺，当我尝试即时搜索时，它会中断并抛出错误。我在没有JSON的情况下写了这个相同的程序，这里是javascript returns blank line at front of result我在那篇文章中建议adeneo重新编程。在我更改代码后，我的页面按预期加载...解决了我在该部分的问题。奇怪的是，它被投了3次：/。对JQuery不感兴趣...不太了解它
答案 0 :(得分：0)
好的....砍伐相当迟钝......哈哈。代码完美无缺。我之前忘了一件简单的事
var d = JSON.parse(create.responseText);
我没有清除数据的div ...我的搜索连接到列表的末尾，远在页面底部附近。我在上面的行之前添加了这段代码...
_("list").innerHTML = '';
现在，当我使用搜索框时，返回数据会清除，只留下我正在寻找的内容。最简单的事情被挫折所忽视：/