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我正在使用PHP（buscador.php）的搜索頁面，我需要的是顯示（總是在同一頁面buscador.php中）員工信息取決於類型ID，因爲我編碼了一個JQuery函數使用$ .ajax爲了通過AJAX和JSON格式獲得數據（數據庫查詢編碼在process.php）
我的問題是，當我定義的ID，然後點擊提交按鈕，它重定向我process.php，而不是告訴我裏面buscador.php特殊區域的信息，該信息顯示process.php是：
{"users":{"status":"OK","0":{"ID":"001","Nombre":"algo","cargo":"algo cargo"},"1":{"ID":"PHP001","Nombre":"Pablo Ernesto Tobar Mayora","cargo":"Web Programmer"},"2":{"ID":"PHP002","Nombre":"Pabletoreto Blogger","cargo":"Blogger Manager"},"3":{"ID":"PHP003","Nombre":"prueba de stored procedure en MySQL","cargo":"Database Administrator"},"4":{"ID":"PHP004","Nombre":"Prueba de funciones en MySQL","cargo":"Database Tester"}}}
這不是結果我'或者你可以看到對數據庫的查詢已經完成，除此之外，信息以JSON格式顯示，但我無法在buscador.php頁面中顯示該信息，請問我是誰做錯了，請，我的整個代碼是：
buscador.php
<html lang="es-ES">
<head>
<meta name="tipo_contenido" content="text/html;" http-equiv="content-type" charset="utf-8">
<link type="text/css" rel="stylesheet" href="content/estilos.css">
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script src="javascript/jquery.js" type="text/javascript"></script>
<title>BUSCADOR</title>
</head>
<form method="post" id="form_id" action="process.php">
<fieldset>
<legend> Buscador Asincrono</legend>
<p>ID a buscar: <input type="text" name="ID_name" id="ID_id"/><div id="estado_id"></div></p>
<p><input type="submit" id="submit_id" value="Buscar"/></p>
<img src="imagenes/cargando.gif" id="LoadingImage" style="display:none" align="center"/>
<div id="ajax_id"><b>Person info will be listed here...</b></div>
<div id="msg">
<table id="infoemp" border="1" style="display:none" align="center">
<thead>
<th>ID</th>
<th>Nombre</th>
<th>Cargo</th>
</thead>
<tbody></tbody>
</table>
</div>
</fieldset>
</form>
</html>
的jquery.js
$(document).ready(function() {
$("#form_id").submit(function(e){
e.preventDefault();
if(validaForm()){
requestInfo();
}
});
});
function validaForm(){
var id_val = $("input#ID_id").val().trim();
//var id_val = id.val().trim();
if((id_val=="") || id_val.replace(/s+/,'') == ''){
alert("Favor ingrese el ID");
$("input#ID_id").addClass("posicionamiento");
$("#ajax_id").html("<div class='error'>Debe especificar el nombre</div>");
return false;
}else{
$("input#ID_id").removeClass("posicionamiento");
$("#div_id").empty();
}
return true;
}
function requestInfo(){
$("#submit_id").hide();
$("#ajax_id").html("");
$("#LoadingImage").show();
$("#ajax_id").html("<div class='cargando'> realizando busqueda</div>");
var url = $("#form_id").attr('action');
var data = $("#form_id").serialize();
var type = $("#form_id").attr('method');
alert(url);
$.ajax({
url:url,
data:data,
type:type,
cache: false,
contentType: "application/x-www-form-urlencoded",
dataType: 'json',
encode: true,
});
.done(function(data) {
if(data.status == "OK"){
$("#submit_id").show();
$("#ajax_id").html("");
$("#LoadingImage").fadeOut();
$("#infoemp").show();
$.each(data.users, function(i,user){
var tblRow =
"<tr>"
+"<td>"+user.ID+"</td>"
+"<td>"+user.Name+"</td>"
+"<td>"+user.cargo+"</td>"
+"</tr>" ;
$(tblRow).appendTo("#infoemp tbody");
});
} else { $("#ajax_id").html(data.status).addClass("cargando"); }
});
.fail(function(jqXHR, textStatus, errorThrown) {
if (console && console.log) {
console.log("La solicitud a fallado: " + textStatus);
}
});
}
process.php
<?php
$bd = "ejemplo";
$server ="localhost";
$user = "root";
$password = "";
if (isset($_POST['Submit']) && isset($_POST['ID_name'])) {
$valor = filter_var($_POST['ID_name'], FILTER_SANITIZE_STRING);
$var = array();
if ($valor == null)
$var["status"]="ERROR";
exit();
} else {
$mysqli = mysqli_connect($server, $user, $password, $bd);
if(! $mysqli) die("Error de conexion ".mysqli_connect_error());
$var["status"]="OK";
$sql = "SELECT * FROM empleado_php";
$result = mysqli_query($mysqli, $sql);
while($obj = mysqli_fetch_assoc($result)) {
$var[] = $obj;
}
$mysqli->close();
}
echo '{"users":'.json_encode($var).'}';
header('Content-type: application/json; charset=utf-8');
?>
爲什麼你這樣做'echo'{「users」：'。json_encode（$ var）。'}'？ – Akar
您已經在阻止默認操作。有沒有錯誤？ –
你好，是的，我已經阻止了默認的動作，我不明白爲什麼它重定向到process.php –