Question

我尝试编写get-methods以将数据插入表中。实际上，一张桌子一切顺利。看一下代码：

<?php
error_reporting(E_ALL & ~E_DEPRECATED);

$db_host = "..."; 
$db_user = "...";
$db_password = "...";
$db_table = "Task"; 

$name = $_GET['Name'];
$groupId = $_GET['GroupId'];
$creatorId = $_GET['CreatorId'];
$comment = $_GET['Comment'];

$db = mysql_connect($db_host, $db_user, $db_password) OR DIE("DB connection fail...");
mysql_select_db("...", $db);
mysql_query("SET NAMES 'utf8'", $db);

$result = mysql_query ("INSERT INTO ".$db_table." (Name, Group_ID, Creator_ID, Comment) VALUES ('$name', '$groupId', '$creatorId', '$comment')");
$id = mysql_insert_id();

header('Content-Type: application/json');


if ($result = 'true'){
    $response = array( 'result' => 'OK', 'id' => $id );
    //setcookie("TaskManagerUser", $id);
    echo json_encode($response);
} else{
    $response = array( 'result' => 'FAIL');
    echo json_encode($response);
}
?>

但是当我尝试插入另一个表＆＃34; Group＆＃34;时，没有任何反应。 mysql_insert_id（）中的id始终为0; 两个表中的主键是AU和唯一

<?php
error_reporting(E_ALL & ~E_DEPRECATED);

$db_host = "..."; 
$db_user = "...";
$db_password = "...";
$db_table = "Group"; 

$name = $_GET['Name'];

$db = mysql_connect($db_host, $db_user, $db_password) OR DIE("DB connection fail...");
mysql_select_db("...", $db);
mysql_query("SET NAMES 'utf8'", $db);

$result = mysql_query("INSERT INTO ".$db_table." (Name) VALUES ('$name')");

$id = mysql_insert_id();

header('Content-Type: application/json');

if ($result = 'true'){
    $response = array( 'result' => 'OK', 'id' => $id );
    //setcookie("TaskManagerUser", $id);
    echo json_encode($response);
} else{
    $response = array( 'result' => 'FAIL');
    echo json_encode($response);
}
?>

Answer 1

对于那些决定创建一个名为＆＃34; Group＆＃34; - 不要这样做！这就是重点。我更改了名称并且有效！

php mysql不插入数据

1 个答案: