我已经从PHP页面发送了json_encoded数据:
$arrayForDataTablePopulation = array();
while (($row = oci_fetch_array($stid, OCI_BOTH)) != false) {
$a = trim($row["IDCONF"]);
$b = trim($row["NOME"]);
$c = trim($row["TIPOLOGIA"]);
$d = trim($row["STATO"]);
$e = trim($row["AMBIENTE"]);
$f = trim($row["AGGREGATORE"]);
$f2 = trim($row["NOMECED"]);
$g = trim($row["MODELLO"]);
$h = trim($row["CONTRATTO"]);
$i = trim($row["VALIDADAL"]);
$l = trim($row["VALIDAAL"]);
if ($count < $NUM){
$arrayForDataTablePopulation[$count] = "[" . $a . "," . $b . "," . $c . "," . $d . "," . $e . "," . $f . "," . $f2 . "," . $g . "," . $h . "," . $i . "," . $l . " ]";
$count++;
} else {
$arrayForDataTablePopulation[$count] = "[" . $a . "," . $b . "," . $c . "," . $d . "," . $e . "," . $f . "," . $f2 . "," . $g . "," . $h . "," . $i . "," . $l . " ]";
}
}
echo json_encode($arrayForDataTablePopulation);
} else {
echo "No data Received. Dimensione array: " . $receivedArrayFormSize;
}
我不知道为什么我的输出不好:
我认为我从服务器对阵列进行了错误编码。 Ajax方面是:
$.ajax({
url: "asset_GestAsset.php",
type: "POST",
data: { "fieldValue" : JSON.stringify(arrForm) },
success: function(data) {
var tableData = JSON.parse(data);
console.log(" Converti in stringhe: " + data.toString() + " !!! Everything ok");
$('#OUTCOME_RESEARCH_TABLE').DataTable({
"aaData":tableData,
"deferLoading": 57,
"deferRender": true,
"scrollY": 150,
"iDisplayLength": 100,
"scrollX": true,
"bRetrieve": true,
"bDestroy": true,
"ordering": false,
"info": true,
"sDom": 'ltipr',
"bDestroy": true
});
etc...
问题是我对所有这些Array / JSON的东西感到困惑,我认为我现在在脑海中非常困惑xD 有人可以帮助我找出解决此问题的方法吗? PHP方面是错误的还是AJAX中的代码?
我想做的是修改服务器阵列,发送回客户端并正确呈现DataTable。
答案 0 :(得分:1)
PHP
您应使用以下代码填充数组:
$arrayForDataTablePopulation[$count] = array($a, $b, $c, $d, $e, $f, $f2, $g, $h, $i, $l);
要打印JSON,请使用以下代码:
header('Content-type: application/json');
echo json_encode($arrayForDataTablePopulation);
JS
在success
处理程序中使用以下代码:
$('#OUTCOME_RESEARCH_TABLE').DataTable({
"dom": 'ltipr',
"data": data,
"pageLength": 100,
"scrollY": 150,
"scrollX": true,
"ordering": false
});