它是一个完美的JSON字符串。
(&P; 这是我的代码,
$("#button").click(function(e)
{
data = {};
$(document).find('[name]').each(function(index , value)
{
name = $(this).attr('name'),
value = $(this).val();
data[name] = value;
});
dataString = JSON.stringify(data);
//console.log(dataString);
$.ajax({
url : "empcontact.php",
type : "post",
data : dataString,
success : function (response) {
console.log(response);
}
})
return false;
});
----------------------------------------- -----------------------------------------
$requestpayload = file_get_contents("php://input");
$array = json_decode($requestpayload , true);
foreach($array as $row)
{
$name = $row['txtname'];
$gender = $row['txtgender'];
$designation = $row['txtdesignation'];
$sql = "INSERT INTO tbl_employee (Name, Gender, Designation) VALUES ('$name' , '$gender' , '$designation')";
mysqli_query($connect, $sql);
}
echo "Employee data submitted successfully";
您通过ajax发送的JSON不是Array类型。 您应该将数组发送到服务器,或者将其视为服务器上的一个简单对象。
如果您的JS代码是正确的,那么您需要修改您的PHP:
$row = json_decode($requestpayload , true);
$name = $row['txtname'];
$gender = $row['txtgender'];
$designation = $row['txtdesignation'];
$sql = "INSERT INTO tbl_employee (Name, Gender, Designation) VALUES ('$name' , '$gender' , '$designation')";
mysqli_query($connect, $sql);