我写的是一个筛选的功能，我用ajax向php中的数据库请求信息，并用JSON传回给javascript，但是输入相关内容时，有的筛选可以从数据库中读出来，有的就不行，想问问怎么解决。

javascript代码：

submitElement.addEventListener('click',function(event){
     if(searchDescription === 1){
         search = filterByTitleElement.value;
     }else if(searchDescription === 2){
         search = filterByDescriptionElement.value;
     }

     //将输入的搜索信息传递给php文件，让其再数据库中找到相应的匹配项。

         var xmlhttp;
         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) {
                 a = eval('(' + xmlhttp.responseText + ')');
                                  tableDisplay.innerHTML = a.length;
                 }
             }
         };
         xmlhttp.open("GET", "SearchDeal.php?inputMessage=" + search + "&searchdescription=" + searchDescription, true);
         xmlhttp.send();
 });

 function pageForDisplay(number){
     var result = "<thead><tr><td>Search Result</td></tr></thead><tr><td><table border=\"1\" id=\"inside\">";
     for(var i = 0; i < 5 ;i++){
         result = result + " <tr><td><p class=\"photoInside\"><a href=\"DetailsPage.php\"><img src=\"travel-images/square-medium/" + a[number + i]['PATH'] +"\"></a>" +
             "<h3><a href=\"DetailsPage.php\">" + a[number + i]['Title'] +"</a></h3>" +
             "<p>" + a[number + i]['Description'] + "</p>" +
             "</p></td></tr>";
     }
      result = result + "</table></td></tr>";
     return result;

php代码：

     $messageForSearch = $_GET["inputMessage"];
$messageForNumber = $_GET["searchdescription"];
$conn = mysqli_connect("localhost","root","","travel");
if($messageForNumber == 1){
    $sql = "SELECT ImageID,Title,Description ,PATH FROM travelimage where Title REGEXP '[$messageForSearch]'";
}else if($messageForNumber == 2){
   $sql = "SELECT ImageID,Title,Description ,PATH FROM travelimage where Description REGEXP '[$messageForSearch]'";
}
$a = mysqli_query($conn,$sql);
$json = mysqli_fetch_all($a,MYSQLI_ASSOC);
//print_r($json);
echo json_encode($json);
}

按照上面的代码，有的时候a.length会改变，但是有时就不会改变（一般是当筛选出来的结果太多时）。然而，如果我将php中的最后一行代码换为print_r($json)时，就会有输出。