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嗨我有一个jqgrid设置与自定义操作按钮,将选定的行发送到数据库。我有一切工作,除了它发布的数据每行7次选择我只希望它发布一次选定的行。请任何和所有帮助表示赞赏。 JQGrid, How to post JSON string to PHP to Process and send to Database?JQGRID,如何将选定的行数据只发布一次到数据库?
$decarr= array(
['id'] =>
518
['name'] =>
'Brochure for amada'
['id_continent'] =>
' Ramon'
['lastvisit'] =>
'5/15/2013'
['cdate'] =>
'5/29/2013'
['ddate'] =>
'5/31/2013'
['email'] =>
'<a href="/media-management/uploads/1368166339.zip">Files</a>'
)
PHP:
//First decode the array
$arr = $_POST["json"];
$decarr = json_decode($arr, true);
$count = count($decarr);
$values = array(); // This will hold our array values so we do one single insert
for ($x=0; $x < $count; $x++){
$newrec = $decarr;
$id = $newrec['id']; $id = mysql_real_escape_string($id);
$name = $newrec['name']; $name = mysql_real_escape_string($name);
$id_continent = $newrec['id_continent']; $id_continent = mysql_real_escape_string($id_continent);
$email = $newrec['email']; $email = mysql_real_escape_string($email);
$lastvisit = $newrec['lastvisit']; $lastvisit = mysql_real_escape_string($lastvisit);
$cdate = $newrec['cdate']; $cdate = mysql_real_escape_string($cdate);
$ddate = $newrec['ddate']; $ddate = mysql_real_escape_string($ddate);
// Create insert array
$values[] = "('".$id."', '".$name."', '".$id_continent."', '".$lastvisit."','".$cdate."','".$ddate."','".$email."')";
}
// Insert the records
$sql = "INSERT INTO finish (id, name, id_continent, lastvisit,cdate,ddate, email)
VALUES ".implode(',', $values);
$result = mysql_query($sql, $con) or die(mysql_error());
?>
[请不要在新代码中使用'mysql_ *'函数](http://stackoverflow.com/q/12859942/1190388)。他们不再被维护,并[正式弃用](https://wiki.php.net/rfc/mysql_deprecation)。看到红色框?改为了解准备好的语句,然后使用[tag:PDO]或[tag:MySQLi]。 – hjpotter92
重复该问题并不能改善其被回答的机会。不过,提炼它确实如此。 – raina77ow
我明白我是一个noob,所以我想抓住php。我会养成这种习惯。关于我的问题,你有什么建议? – NewHistoricForm