在我的Ajax调用成功后,我想用我的php部分返回的新值重新加载我的bootstrap表。
<table id="gelirtableid" data-toggle="table" data-url="gelir-getdata.php" data-classes="table table-hover" data-striped="true"
data-pagination="true" data-page-list=[20, 40, 75, 100] data-search="true">
<thead>
<tr >
<th data-sortable="true" data-field="tarih">Tarih</th>
<th data-sortable="true" data-field="Toplam">Toplam</th>
</tr>
</thead>
</table>
我的php脚本从mysql中获取数据
<?php
include "dbcon.php";
if($_POST["gelirtablosecimi"]){
$gelirtabloadi = $_POST["gelirtablosecimi"];
$_SESSION["gelirtabloadi"] = $gelirtabloadi;
}
$gelirtabloadi = $_SESSION["gelirtabloadi"];
$gelirgunluktoplam = $db->prepare("select tarih, hasilat + visa + butce_ici + hisse_satis + sosyal_konut + elektrik + haberlesme + iller_bank + diger AS Toplam from $gelirtabloadi");
$gelirgunluktoplam->execute();
$results = $gelirgunluktoplam->fetchAll(PDO::FETCH_ASSOC);
$json = json_encode($results);
echo $json;
?>
和我的Ajax电话
$("#gelirgetir").click(function() {
var gelirtablosecimi = $("#select1").val();
if (gelirtablosecimi) {
$.ajax({
type: "POST",
url: "gelir-getdata.php",
data: {
"gelirtablosecimi": gelirtablosecimi
},
success: function(result) {
notifyUser('success', 'Başarılı!', 'Tablo başarıyla güncellendi');
location.reload();
},
error: function(result) {
notifyUser('error', 'Hata', 'error');
}
});
} else {
notifyUser('info', 'Dikkat', 'Tablo seçimi yapmadınız!');
}
});
我是Ajax调用的新手,可能问题在于我的ajax部分。如你所见,我在电话会议成功后正在与location.reload();
打交道。我尝试重装.container
和#gelirtableid
,但没有任何对我有用。基本上,当我按下#gelirgetir
按钮时,它会更新我的一个会话值,而我的表依赖于该会话值。会话值更改后,如果我重新加载页面,将显示新值,但我必须强制刷新页面。我只想刷新表格。有什么建议吗?
替换您的表代码,以便您有一个表体。我们稍后会替换它。
<table id="gelirtableid" data-toggle="table" data-url="gelir-getdata.php" data-classes="table table-hover"
data-striped="true"
data-pagination="true" data-page-list=[20, 40, 75, 100] data-search="true">
<thead>
<tr>
<th data-sortable="true" data-field="tarih">Tarih</th>
<th data-sortable="true" data-field="Toplam">Toplam</th>
</tr>
</thead>
<tbody>
</tbody>
</table>
在您的函数中,您需要通过jQuery中的parseJSON函数从响应中提取数据。然后,您需要为桌面构建内容。这是一个例子。
$("#gelirgetir").click(function () {
var gelirtablosecimi = $("#select1").val();
if (gelirtablosecimi) {
$.ajax({
type: "POST",
url: "gelir-getdata.php",
data: {
"gelirtablosecimi": gelirtablosecimi
},
success: function (result) {
notifyUser('success', 'Başarılı!', 'Tablo başarıyla güncellendi');
var content = '';
$.each(result, function(i, data) {
content += '<tr>';
content += '<td>'+result[i].tarih+'</td>';
content += '<td>'+result[i].Toplam+'</td>';
content += '</tr>';
});
$('#gelirtableid tbody').html(content);
},
error: function (result) {
notifyUser('error', 'Hata', 'error');
}
});
} else {
notifyUser('info', 'Dikkat', 'Tablo seçimi yapmadınız!');
}
});
当我从上面的解决方案得到的错误是关于JSON.parse()
。 $.each
希望解析JSON数据。以下代码就像魅力一样。
<table id="gelirtableid" data-toggle="table" data-url="gelir-getdata.php" data-classes="table table-hover" data-striped="true"
data-pagination="true" data-page-list=[20, 40, 75, 100] data-search="true">
<thead>
<tr >
<th data-sortable="true" data-field="tarih">Tarih</th>
<th data-sortable="true" data-field="Toplam">Toplam</th>
</tr>
</thead>
</table>
Ajax调用:
$("#gelirgetir").click(function() {
var gelirtablosecimi = $("#select1").val();
if (gelirtablosecimi) {
$.ajax({
type: "POST",
url: "gelir-getdata.php",
data: {
"gelirtablosecimi": gelirtablosecimi
},
success: function(result) {
var content = '';
var obj = JSON.parse(result);
$.each(obj, function(i, data) {
content += '<tr>';
content += '<td>'+obj[i].tarih+'</td>';
content += '<td>'+obj[i].Toplam+'</td>';
content += '</tr>';
});
$('#gelirtableid tbody').html(content);
notifyUser('success', 'Başarılı!', 'Tablo başarıyla güncellendi');
},
error: function(result) {
notifyUser('error', 'Hata', 'error');
}
});
} else {
notifyUser('info', 'Dikkat', 'Tablo seçimi yapmadınız!');
}
});