我想将login.php变量$user_key和$user_id的值传递给我使用了会话的另一个文件statusdata.php。在statusdata.php中，我想在statusdata.php文件的sql查询中使用这些会话变量值。
要在login.php中实现此目的，我先将变量$user_key和$user_id的值分别传递给会话变量$_SESSION["key"]和$_SESSION["id"]，然后再传递给{{1} }我将会话变量称为变量，并在statusdata.php
$user_key和$user_id中传递它们的值
现在的问题是，在SQL查询中使用变量statusdata.php和$user_key时，它没有返回正确的输出，但是在echo中使用了相同的变量，却给出了适当的值，这意味着会话可以正常工作。回显变量，但在SQL中不起作用。我也尝试过直接传递会话变量，但在echo中却发生了同样的事情，但在SQL中却没有。
$user_id
login.php <?php
// Start the session
session_start();
require "conn.php";
$user_key = '8C9333343C6C4222418EDB1D7C9F84D051610526085960A1732C7C3D763FFF64EC7F5220998434C896DDA243AE777D0FB213F36B9B19F7E4A244D5C993B8DFED';
$user_id = '1997';
$mysql_qry = "select * from applications where application_key = '".$user_key."' and application_id like '".$user_id."';";
$result = mysqli_query($conn, $mysql_qry);
if (mysqli_num_rows($result) > 0){
$_SESSION["key"] = ".$user_key.";
$_SESSION["id"] = ".$user_id.";
echo "Login Success";
}
else {
echo "Login Not Success";
}
?>
statusdata.php 输出<?php
// Start the session
session_start();
require "conn.php";
$user_key = "-1";
if (isset($_SESSION["key"])) {
$user_key = $_SESSION["key"];
}
$user_id = "-1";
if (isset($_SESSION["id"])) {
$user_id = $_SESSION["id"];
}
//creating a query
$stmt = $conn->prepare("SELECT applications.application_id, applications.applicant_name, applications.applicant_aadhaar, applications.applicant_phone, applications.subject, applications.date, applications.description, applications.chairperson_remark, applications.status, officer_accounts.name_of_officer, applications.officer_remark, applications.last_update_on
FROM applications INNER JOIN officer_accounts ON applications.account_id = officer_accounts.account_id
WHERE applications.application_id = '".$user_id."' AND applications.application_key = '".$user_key."';");
//executing the query
$stmt->execute();
//binding results to the query
$stmt->bind_result($id, $name, $aadhaar, $phone, $subject, $date, $description, $chairperson, $status, $officername, $officerremark, $lastupdate);
$applications = array();
//traversing through all the result
while($stmt->fetch()){
$temp = array();
$temp['applications.application_id'] = $id;
$temp['applications.applicant_name'] = $name;
$temp['applications.applicant_aadhaar'] = $aadhaar;
$temp['applications.applicant_phone'] = $phone;
$temp['applications.subject'] = $subject;
$temp['applications.date'] = $date;
$temp['applications.description'] = $description;
$temp['applications.chairperson_remark'] = $chairperson;
$temp['applications.status'] = $status;
$temp['officer_accounts.name_of_officer'] = $officername;
$temp['applications.officer_remark'] = $officerremark;
$temp['applications.last_update_on'] = $lastupdate;
array_push($applications, $temp);
}
//displaying the result in json format
echo json_encode($applications);
// Echo session variables that were set on previous page
echo "<br>key is " . $user_key . ".<br>";
echo "id is " . $user_id . ".";
?>
login.php 输出Login Success
statusdata.php 我想从[]
key is .8C9333343C6C4222418EDB1D7C9F84D051610526085960A1732C7C3D763FFF64EC7F5220998434C896DDA243AE777D0FB213F36B9B19F7E4A244D5C993B8DFED..
id is .1997..
获得的输出（如果我在变量statusdata.php和$user_key中使用直接值，而不是login.php中的会话变量，我会得到它）
$user_id 注意：我正在以JSON格式获取[{"applications.application_id":1997,"applications.applicant_name":"Tanishq","applications.applicant_aadhaar":"987654321","applications.applicant_phone":"123456789","applications.subject":"asdnjsnadnksncdnsjnvsavasdnjsnadnksncdnsjnvsav","applications.date":"2018-07-02 09:11:47","applications.description":"asdnjsnadnksncdnsjnvsavasdnjsnadnksncdnsjnvsavasdnjsnadnksncdnsjnvsavasdnjsnadnksncdnsjnvsavasdnjsnadnksncdnsjnvsav","applications.chairperson_remark":"asdnjsnadnksncdnsjnvsavasdnjsnadnksncdnsjnvsav","applications.status":1,"officer_accounts.name_of_officer":"Chayan Bansal","applications.officer_remark":"asdnjsnadnksncdnsjnvsavasdnjsnadnksncdnsjnvsav","applications.last_update_on":"2018-07-22 09:14:25"}]
key is 8C9333343C6C4222418EDB1D7C9F84D051610526085960A1732C7C3D763FFF64EC7F5220998434C896DDA243AE777D0FB213F36B9B19F7E4A244D5C993B8DFED.
id is 1997.
SQL查询的输出，最后在Android中将其提取。
请帮助我，我已经尝试了其他类似问题所建议的一切，但没有帮助
答案 0 :(得分：0)
看起来更像是错别字。您在login.php中有此内容：
$_SESSION["key"] = ".$user_key.";
$_SESSION["id"] = ".$user_id.";
通过添加不必要的点来污染您的原始值。只需简单分配即可：
$_SESSION["key"] = $user_key;
$_SESSION["id"] = $user_id;
它应该会开始更好地工作。
旁注，您正在使用prepare，但没有使用bind_param。将您的SQL prepare更改为以下内容（请注意?占位符），然后添加bind_param：
$stmt = $conn->prepare("SELECT applications.application_id, applications.applicant_name, applications.applicant_aadhaar, applications.applicant_phone, applications.subject, applications.date, applications.description, applications.chairperson_remark, applications.status, officer_accounts.name_of_officer, applications.officer_remark, applications.last_update_on
FROM applications INNER JOIN officer_accounts ON applications.account_id = officer_accounts.account_id
WHERE applications.application_id = ? AND applications.application_key = ?;");
$stmt->bind_param('ss',$user_id,$user_key);