我向php文件发送一个jQuery请求(包含一个business_id),以检索数据库中的所有值,以填充我的表单中与该id对应的字段和选择。但是,我如何能够从数据库中分块检索响应?这样我就可以提供带有数据库值的表单中的字段和选择。我的javascript函数如下所示:
businessselect: function(){
$('#busselect').change(function() {
opt = $(this).val();
if (opt=="new_bus") {
location.reload();
}
else
{
businessid = $(this).children(":selected").attr("id");
$.ajax({
url : "businessdata.php",
method : "post",
data : "business_id="+businessid,
success: function(response) {
$("#uitgevoerd_door_naam").val(response);
}
});
}
});
},
我的businessdata.php如下所示:
<?php
$mysqli = new mysqli("localhost", "root", "", "brandveiligheid");
if ($mysqli->connect_errno) {
printf("Connect failed: %s'n", $mysqli->connect_error);
exit();
}
if($_POST)
{
$result = $mysqli->query("SELECT * from form WHERE ID ='$_POST[business_id]'");
while ($row = $result->fetch_assoc()) {
echo $row['uitgevoerd_door_naam'];
echo $row['hoev_gev_stof_score'];
}
}
mysqli_close($mysqli);
?>
我想要实现的是:
$("#uitgevoerd_door_naam").val() == $row['uitgevoerd_door_naam'];
$("#hoev_gev_stof_score").val() == $row['hoev_gev_stof_score'];
等等。。。。。
修复:
使用json编码:
功能:
businessselect: function(){
$('#busselect').change(function() {
opt = $(this).val();
if (opt=="new_bus") {
location.reload();
}
else
{
businessid = $(this).children(":selected").attr("id");
$.ajax({
url : "businessdata.php",
method : "post",
dataType: "json",
data : "business_id="+businessid,
success: function(response) {
$("#uitgevoerd_door_naam").val(response.a);
$("#riskpot_scorefield3").val(response.b);
}
});
}
});
},
php文件:
<?php
$mysqli = new mysqli("localhost", "root", "", "brandveiligheid");
if ($mysqli->connect_errno) {
printf("Connect failed: %s'n", $mysqli->connect_error);
exit();
}
if($_POST)
{
$result = $mysqli->query("SELECT * from form WHERE ID = '$_POST[business_id]'");
while ($row = $result->fetch_assoc()) {
echo json_encode(array("a" => $row['uitgevoerd_door_naam'], "b" => $row['hoev_gev_stof_score']));
}
}
mysqli_close($mysqli);
?>