我是PHP和SQL编码的新手,一直在尝试将数据发布到我的数据库中,但当使用Postman检查响应时,我得到了else响应,我真的不知道为什么,因为我的发布数据正是所需的。
非常感谢你对我的帮助!
根据请求,以下是代码
<?php
require("../database_android.php");
if(isset($_POST["package_name"]) && $_POST["package_name"] != "") {
$package_name = mysql_real_escape_string($_POST["package_name"]);
} else {
echo "package_name missing";
return;
}
if(isset($_POST["app_name"]) && $_POST["app_name"] != "") {
$app_name = mysql_real_escape_string($_POST["app_name"]);
} else {
echo "app_name missing";
return;
}
if(isset($_POST["activity"]) && $_POST["activity"] != "") {
$activity = mysql_real_escape_string($_POST["activity"]);
} else {
echo "activity missing";
return;
}
if(isset($_POST["activity_name"]) && $_POST["activity_name"] != "") {
$activity_name = mysql_real_escape_string($_POST["activity_name"]);
} else {
echo "activity_name missing";
return;
}
if(isset($_POST["component_info"]) && $_POST["component_info"] != "") {
$component_info = mysql_real_escape_string($_POST["component_info"]);
} else {
echo "component_info missing";
return;
}
$application_id = already_exists_application($package_name);
if($application_id < 0) {
mysql_query("INSERT INTO androidactivities_application (package_name, app_name) VALUES ('". $package_name ."', '". $app_name ."')");
$application_id = already_exists_application($package_name);
}
$componentinfo_id = already_exists_componentinfo($application_id, $component_info);
if($componentinfo_id < 0 && $application_id > 0) {
mysql_query("INSERT INTO androidactivities_componentinfo (application_id, activity, activity_name, component_info) VALUES ('". $application_id ."', '". $activity ."', '". $activity_name ."', '". $component_info ."')");
echo "successful";
} else {
echo "exists";
}
?>
<?php
function already_exists_application($package_name) {
$result = mysql_query("SELECT * from androidactivities_application WHERE package_name = '".$package_name."'");
if($row = mysql_fetch_object($result)) {
return $row -> application_id;
}
else {
return -1;
}
}
function already_exists_componentinfo($application_id, $component_info) {
$result = mysql_query("SELECT * from androidactivities_componentinfo WHERE application_id = '".$application_id."' AND component_info = '".$component_info."'");
if($row = mysql_fetch_object($result)) {
return $row -> componeninfo_id;
}
else {
return -1;
}
}
?>
发布于 2021-01-13 16:05:59
请在Body标签中以表单数据的形式在postman中发布您的数据。您发送它的方式如下所示的ApiURL?param1=value1¶m2=value2
查看此图像:https://pasteboard.co/JJt9RB4.png
https://stackoverflow.com/questions/65705263
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