I have written code for button to send a value to php file using ajax as show below :
我已经为button编写了使用ajax向php文件发送值的代码,如下所示:
<script>
$(document).ready(function(){
$(".open-AddBookDialog").click(function(){
var packageid = $(this).attr("data-id");
var dataString = 'packageid='+ packageid;
$.ajax({
type: "POST",
url: "data1.php",
data: dataString,
cache: false,
success: function(result1){
var packagetype = package_type;
alert(packagetype);
}
});
});
});
</script>
This is the ajax code for the button which i have written. My button code is :
这是我编写的按钮的ajax代码。我的按钮代码是:
<a data-toggle="modal" id="custId" data-name="<?php echo $arr["package_name"]; ?>" data-id="<?php echo $arr["id"]; ?>" href="#myModal" class="open-AddBookDialog btn btn-submit" OnClick='change(this);'>Book / Enquiry / Pay</a>
When clicking this button in href, I want to send a id value to a php file using ajax.
在href中单击此按钮时,我希望使用ajax向php文件发送id值。
data1.php
data1.php
<?php
include('database.php');
$id=$_POST['packageid'];
$query1 = mysqli_query($con,"select * from ayurveda_packagetypes where package_id = '$id'");
$arr = mysqli_fetch_array($query1);
echo $arr['package_type'];
echo $arr['package_price'];
mysqli_close($con);
?>
using the id value, I want to pick multiple rows of package type and package price from the database having the same id value.After picking the multiple rows of these values i want to send it to the main php file and display all the values of all the rows picked from the database.
使用id值,我想从具有相同id值的数据库中选择包类型和包价格的多行。在选择这些值的多行之后,我希望将其发送到主php文件,并显示从数据库中选择的所有行的所有值。
Can anyone suggest how to do this ?
谁能告诉我该怎么做吗?
2 个解决方案
#1
2
<script>
$(document).ready(function()
{
$(".open-AddBookDialog").click(function()
{
var packageid = $(this).attr("data-id");
var dataString = 'packageid='+ packageid;
$.ajax(
{
type: "POST",
url: "data1.php",
data: dataString,
cache: false,
success: function(result)
{
resultJson=jQuery.parseJSON(result);
$.each(resultJson.packageDetails, function(i, item) {
var packagetype = item.package_type;
var package_price = item.package_price;
alert("packagetype :- "+packagetype+" ----- package_price :-"+package_price);
});
}
});
});
});
</script>
<?php
include('database.php');
$id=$_POST['packageid'];
$query1 = mysqli_query($con,"select * from ayurveda_packagetypes where package_id = '$id'");
//$arr = mysqli_fetch_array($query1);
while( $strPackageResult=mysqli_fetch_array($query1,MYSQLI_ASSOC) )
{ $ArrPackage[]=$strPackageResult; }
if( isset($strPackageResult) ){ unset($strPackageResult); }
mysqli_free_result($query1);
if( isset($query1) ){ unset($query1); }
$myResultPackageArray=array();
if( isset($ArrPackage) && is_array($ArrPackage) && count($ArrPackage)>0 ) {
$tempArray=array();
foreach( $ArrPackage as $tempPackage )
{
$tempArray['package_type']=$tempPackage['$tempPackage'];
$tempArray['package_price']=$tempPackage['package_price'];
$myResultPackageArray['packageDetails'][] =$tempArray;
}
}
mysqli_close($con);
echo json_encode($myResultPackageArray);
?>
#2
1
There are some basic things you should know first then you can easily rectify your errors.
有一些基本的东西你应该首先知道,然后你可以很容易地纠正你的错误。
-
调试javascript
-
准备好的语句
-
PHP的错误处理
This is not you have asked but as a programmer i will suggest you to go through it.
这不是你所要求的,但作为一个程序员,我建议你去做。
As going through your code
通过你的代码
var packagetype = package_type;
package_type
is undefined. If you are using chrome inspect element and check the console you will see the error.
package_type是未定义的。如果您正在使用chrome inspect元素并检查控制台,您将看到错误。
Hope this will work.
希望这将工作。
#1
2
<script>
$(document).ready(function()
{
$(".open-AddBookDialog").click(function()
{
var packageid = $(this).attr("data-id");
var dataString = 'packageid='+ packageid;
$.ajax(
{
type: "POST",
url: "data1.php",
data: dataString,
cache: false,
success: function(result)
{
resultJson=jQuery.parseJSON(result);
$.each(resultJson.packageDetails, function(i, item) {
var packagetype = item.package_type;
var package_price = item.package_price;
alert("packagetype :- "+packagetype+" ----- package_price :-"+package_price);
});
}
});
});
});
</script>
<?php
include('database.php');
$id=$_POST['packageid'];
$query1 = mysqli_query($con,"select * from ayurveda_packagetypes where package_id = '$id'");
//$arr = mysqli_fetch_array($query1);
while( $strPackageResult=mysqli_fetch_array($query1,MYSQLI_ASSOC) )
{ $ArrPackage[]=$strPackageResult; }
if( isset($strPackageResult) ){ unset($strPackageResult); }
mysqli_free_result($query1);
if( isset($query1) ){ unset($query1); }
$myResultPackageArray=array();
if( isset($ArrPackage) && is_array($ArrPackage) && count($ArrPackage)>0 ) {
$tempArray=array();
foreach( $ArrPackage as $tempPackage )
{
$tempArray['package_type']=$tempPackage['$tempPackage'];
$tempArray['package_price']=$tempPackage['package_price'];
$myResultPackageArray['packageDetails'][] =$tempArray;
}
}
mysqli_close($con);
echo json_encode($myResultPackageArray);
?>
#2
1
There are some basic things you should know first then you can easily rectify your errors.
有一些基本的东西你应该首先知道,然后你可以很容易地纠正你的错误。
-
调试javascript
-
准备好的语句
-
PHP的错误处理
This is not you have asked but as a programmer i will suggest you to go through it.
这不是你所要求的,但作为一个程序员,我建议你去做。
As going through your code
通过你的代码
var packagetype = package_type;
package_type
is undefined. If you are using chrome inspect element and check the console you will see the error.
package_type是未定义的。如果您正在使用chrome inspect元素并检查控制台,您将看到错误。
Hope this will work.
希望这将工作。