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我無法通過ajax發送序列化表單到php文件。我可以看到客戶端的字符串,但在服務器端我收到一個空數組。 我試圖將表單數據保存到數據庫中,但我似乎無法找到一種方法來分隔每個輸入,並在我發送了ajax後在我的php文件中顯示它。序列化表單不發送ajax
的JavaScript
$(function() {
//twitter bootstrap script
$("button#guardar").click(function(e) {
//var info = $('#myform').serialize();
var info = $('form.contact').serialize();
$.ajax({
type: "POST",
url: "solicitudesProc.php",
data: info,
success: function(data) {
alert(info);
window.location.href = "solicitudesProc.php";
//window.location.reload();
$("#modalnuevo").modal('hide');
},
error: function(data) {
alert("failure");
}
});
});
});
<form class="contact" id="myform" method="post" name='alta'>
<div class="modal-body">
<div class="row">
<div class="col-md-2">
<label>Solicitante</label>
<input type="text" class="form-control pull-right" name='solicitante' maxlength="20" required />
</div>
<div class="col-md-2">
<label>Fecha Emision</label>
<input type="text" class="form-control pull-right" name='fechaEmision' maxlength="20" />
</div>
</div>
<div class="row">
<div class="col-md-2">
<label>Area Solicitante</label>
<input type="text" class="form-control pull-right" name='area' maxlength="20" />
</div>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Cerrar</button>
<button type="submit" id="guardar" name='guardar' class="btn btn-danger pull-right" value="guardar">Generar</button>
</div>
</form>
服務器端solicitudesProc.php
<?php $info = $_POST;
echo $_POST["solicitante"]; print_r($_POST); ?>
你爲什麼要改變位置? 'window.location.href =「solicitudesProc.php」;'將使用GET重新加載頁面並且沒有數據 – mplungjan
https://stackoverflow.com/questions/18866571/receive-json-post-with-php – chiliNUT
您還需要取消按鈕的默認操作:提交表單。 – jeroen