I want to send the result of an SQL query which is a PHP variable to the index file. Here's from the PHP I want to send from:
我想将SQL查询的结果作为PHP变量发送到索引文件。这是我想发送的PHP:
$result = mysqli_query($conn, $sql);
$remaining = array();
while($row = mysqli_fetch_array($result)){
array_push($remaining, $row['_id']);
}
echo json_encode($remaining);
?>
<script>
$.ajax({
url: "index.php",
data: {result: result},
dataType: "json",
success: function(result){
}});
</script>
Here is where I want to use the code:
这是我想要使用代码的地方:
$var_value = $_GET['remaining'];
$values = implode(", ", $var_value);
$sql = "SELECT *
FROM `species`
WHERE `_id` IN (".$valus.") ";
$results = mysqli_query($conn, $sql);
I have tried many methods, running a second ajax call within one already running since the $remaining variable is sent back to a javascript using ajax. $_cookies doesn't work for my webserver. So this seems the best solution. Can anyone see what is going wrong?
我已经尝试了很多方法,在已经运行的一个中运行第二个ajax调用,因为$ remaining变量使用ajax发送回javascript。 $ _cookies对我的网络服务器不起作用。所以这似乎是最好的解决方案。任何人都可以看到出了什么问题?
1 个解决方案
#1
As reply to your comment: easy:
回复你的评论:容易:
session_start(); // important!!
if ( !isset($_SESSION['remaining']) )
{
$result = mysqli_query($conn, $sql);
$remaining = array();
while($row = mysqli_fetch_array($result)){
array_push($remaining, $row['_id']);
}
$_SESSION['remaining'] = $remaining;
}
else
{
$values = implode(",", $_SESSION['remaining']);
/// $_SESSION['remaining'] = null - depends on logic.
$sql = "SELECT *
FROM `species`
WHERE `_id` IN (".$valus.") ";
$results = mysqli_query($conn, $sql);
}
#1
As reply to your comment: easy:
回复你的评论:容易:
session_start(); // important!!
if ( !isset($_SESSION['remaining']) )
{
$result = mysqli_query($conn, $sql);
$remaining = array();
while($row = mysqli_fetch_array($result)){
array_push($remaining, $row['_id']);
}
$_SESSION['remaining'] = $remaining;
}
else
{
$values = implode(",", $_SESSION['remaining']);
/// $_SESSION['remaining'] = null - depends on logic.
$sql = "SELECT *
FROM `species`
WHERE `_id` IN (".$valus.") ";
$results = mysqli_query($conn, $sql);
}