我试图在单个ajax请求中将整个(网格)记录范围从存储发送到服务器,并将所有记录保存在另一个表中。
但是,它似乎缺少一些细节。
在客户端,console.log(记录)显示所有要发送的对象。
在客户端,回显“ console.log(“ PHP:”。$ data。'“)'出口;返回console.log(“ PHP:”);
扩展JS 5
newList: function (button, e, eOpts) {
var grid = this.lookupReference('gridRef');
var store = grid.getStore(),
// http://stackoverflow.com/questions/37663867/what-is-the-best-way-to-create-a-list-from-the-rows-of-a-grid
//get whole range of records from the store
var records = store.getRange().map(function(record) { return record.getData() });
//console.log(records); // OK
Ext.Ajax.request({
url: 'php/newList.php?',
method: 'POST',
//jsonData: records,
params:{records:Ext.encode(records)} //EDITED
success: function(conn, response, options, eOpts) {
console.log('Success');
},
failure: function(conn, response, options, eOpts) {
console.log('Error')
}
});
的PHP
$records = stripslashes($_POST['records']);
$data = json_decode($records, true);
//var_dump($_POST);
// echo 'console.log("PHP: '.$data.'")', exit;
foreach($data as $row) {
$item[] = $row['item'];
}
for($i=0; $i<count($data);$i++){
$sqlQuery = "INSERT INTO tab_list (item) VALUES (?)";
}
if($statement = $conexao->prepare($sqlQuery)){
$statement->bind_param("s", $item[$i]);
$statement->execute();
$success= true;
}else{
$erro = $conexao->error;
$success = false;
}
$sucess = array("success" => mysqli_errno($conexao) == 0);
echo json_encode(array(
"success" => $sucess
));
$statement->close();
$conexao->close();
编辑:
我在ajax请求
jsonData: records
中更改了params:{records:Ext.encode(records)}
在PHP代码上使用
var_dump($_POST);
时,我在DevTools上收到以下响应array(10) {
[0]=>
array(5) {
["item"]=>
string(9) "item five"
["id"]=>
string(22) "MyList.listDataModel-1"
}
[1]=>
array(5) {
["item"]=>
string(10) "item seven"
["id"]=>
string(22) "MyList.listDataModel-2"
}
[2]=>
array(5) {
["item"]=>
string(9) "item four"
["id"]=>
string(22) "MyList.listDataModel-3"
}
...
但是,我仍然无法在表中插入这些数据并创建新列表。现在问题出在PHP代码中。
PHP解决方案:
$records = stripslashes($_POST['records']);
$data = json_decode($records, true);
//var_dump($_POST);
// echo 'console.log("PHP: '.$data.'")', exit;
foreach($data as $row) {
$item[] = $row['item'];
}
for($i=0; $i<count($data);$i++){
$sqlQuery = "INSERT INTO tab_list (item) VALUES (?)";
//within for
if($statement = $conexao->prepare($sqlQuery)){
$statement->bind_param("s", $item[$i]);
$statement->execute();
$success= true;
}else{
$erro = $conexao->error;
$success = false;
}
}
$sucess = array("success" => mysqli_errno($conexao) == 0);
echo json_encode(array(
"success" => $sucess
));
$statement->close();
$conexao->close();
最佳答案
$item
是否在该foreach循环之前初始化?看起来您希望if($statement = $conexao->prepare(...
行位于for
(或可能是foreach
)循环内。
编辑问题后,请考虑使用单个INSERT
语句替代解决方案:
$total = count($item);
$sqlQuery = 'INSERT INTO tab_list (item) VALUES ' . rtrim(str_repeat('(?),', $total), ',');
$statement = $conexao->prepare($sqlQuery);
if ($statement)
{
for ($i = 0; $i < $total; $i++)
{
$statement->bind_param('s', $item[$i]);
}
$statement->execute();
$success = true;
}
else
{
$erro = $conexao->error;
$success = false;
}