PHP / JAVASCRIPT / SQL无法显示表

Question

我尝试创建一个查询系统，将系统中的所有用户帐户显示给系统管理员，以便稍后我可以添加一个功能来编辑它们等等。我已经创建了一个PHP文件找到结果并创建一个表，然后回显到javascript并显示。但是，系统仅显示0而不显示表。我只能假设这意味着没有找到任何东西，但我的系统意味着要抓住这一点，我90％肯定它应该找到一些东西。

的Javascript

function UserSearch(){
var Search = $("#txtUserSearch" ).val();
var Type = $("#sltQuery" ).val()
$.post('../functions/php/fncusersearch.php', {Search: Search, Type: Type}, function(data) {
    if (data == 1){
        $('#divSearchResults').html('<p class="text-center text-danger bg-danger" id="pUPInc">No Accounts Matched your Search!</p>');
    }
    else if (data == 3){
        $('#divSearchResults').html('<p class="text-center text-danger bg-danger" id="pUPInc">Database not found! Please try again later.</p>');
    }
    else{
        $('#divSearchResults').html(data);
    }
});

}

PHP

//Retrieves variables from Javascript.
$Search = $_POST["Search"];
$Type = $_POST["Type"];

if ($Type == "Registration Date"){
    $Type = "joined";
}
else if ($Type == "Account Rank"){
    $Type = "rank";
}

include "db/openlogindb.php";
if($DBError == true){
    $data = 3;
}
else{

    $UserSearch = "SELECT username, surname, forename, joined, rank FROM users WHERE ".$Type." LIKE '%".$Search."%'";

    $results = mysqli_query($conn, $UserSearch);

    if(mysqli_num_rows($results) == 0){
        $data = 1;
    }
    else{
        $data = '';
        $data += '<table class="table table-striped">';

        while($row = mysqli_fetch_assoc($results)){
            $data += "<tr><td>".$row['surname']."</td><td>".$row['forename']."</td><td>".$row['username']."</td><td>".$row['joined']."</td><td>".$row['rank']."</td></tr>";
        }

        $data += "</table>";
    }
}

include "db/closelogindb.php";

echo $data;
?>

http://thomas-smyth.co.uk/admin/manageusers.php

Answer 1

您使用错误的运算符进行连接

$data += '<table class="table table-striped">';

用。替换+符号。

例如

$data .= '<table class="table table-striped">';