我试图实现多个if条件来检查提案的状态,并且只应执行特定代码(如果状态代码设置为1或5)。PHP IF/ELSE条件不能按预期工作
由于某种原因,我在实施这方面遇到了困难。目前代码中的逻辑是,如果提案状态不匹配1或5,则返回消息,否则执行下一个查询。当我只指定一个数字,即(1或5)时,它会正常工作。
我与若和其他条件,在这部分面临着另一个问题:
if ($count == 1) {
$feedback = '<p class="text-danger"> You have already accepted an application. You cannot accept or apply for any others. If this is a mistake then please contact the supervisor concerned directly.</p>';
}
if ($count < 1) {
$status = $db_conx->prepare ("SELECT status_code FROM record WHERE student_record_id = :user_record_id AND proposal_id = :proposal");
$status->bindParam(':user_record_id', $user_record_id, PDO::PARAM_STR);
$status->bindParam(':proposal', $proposal, PDO::PARAM_STR);
$status->execute();
$proposalstatus = $status->fetchColumn();
if($proposalstatus != 1)
{
//echo $proposalstatus;
$feedback = '<p class="text-danger">The proposal is not at a status where it can be accepted</p>';
}
}
else {
当我单独运行每个部分,但是当我试图把它一起在if语句中失败,并没有按这个工程根本不检查这些条件,只是完成更新数据库并显示成功消息的任务。
完整的代码是在这里:
try
{
$db_conx = new PDO("mysql:host=$mysql_hostname;dbname=$mysql_dbname", $mysql_username, $mysql_password);
$db_conx->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$username = $_SESSION['username'];
$sql = $db_conx->prepare("SELECT username, user_record_id FROM login_details
WHERE username = :username");
$sql->bindParam(':username', $username, PDO::PARAM_STR);
$sql->execute();
$user_record_id = $sql->fetchColumn(1);
$proposal = $_POST['proposal_id'];
$acceptCheck = $db_conx->prepare ("SELECT * FROM record WHERE student_record_id = :user_record_id AND status_code = 3");
$acceptCheck->bindParam(':user_record_id', $user_record_id, PDO::PARAM_STR);
$acceptCheck->execute();
$count = $acceptCheck->rowCount();
if ($count == 1) {
$feedback = '<p class="text-danger"> You have already accepted an application. You cannot accept or apply for any others. If this is a mistake then please contact the supervisor concerned directly.</p>';
}
if ($count < 1) {
$status = $db_conx->prepare ("SELECT status_code FROM record WHERE student_record_id = :user_record_id AND proposal_id = :proposal");
$status->bindParam(':user_record_id', $user_record_id, PDO::PARAM_STR);
$status->bindParam(':proposal', $proposal, PDO::PARAM_STR);
$status->execute();
$proposalstatus = $status->fetchColumn();
if($proposalstatus != 1 || 5) //status must be either 'Approved' code 1 or 'Held' code 5
{
//echo $proposalstatus;
$feedback = '<p class="text-danger">The proposal is not at a status where it can be accepted</p>';
}
}
else {
//Update all application records to 'Not available' when a proposal has been accepted
$updateOtherRecords = $db_conx->prepare("UPDATE record SET status_code = 8, last_updated = now()
WHERE proposal_id = :proposal");
$updateOtherRecords->bindParam(':proposal', $proposal, PDO::PARAM_STR);
$updateOtherRecords->execute();
//Update other applicationa for the user concerned to 'Rejected'
$updateUserRecord = $db_conx->prepare("UPDATE record SET status_code = 7, last_updated = now()
WHERE student_record_id = :user_record_id");
$updateUserRecord->bindParam(':user_record_id', $user_record_id, PDO::PARAM_STR);
$updateUserRecord->execute();
//Update the proposal concerned and assign it to the user
$update = $db_conx->prepare("UPDATE record SET status_code = 3, last_updated = now()
WHERE proposal_id = :proposal AND student_record_id = :user_record_id");
$update->bindParam(':user_record_id', $user_record_id, PDO::PARAM_STR);
$update->bindParam(':proposal', $proposal, PDO::PARAM_STR);
$update->execute();
$feedback = '<p class="text-success"> The proposal has been successfully accepted <span class="glyphicon glyphicon-ok"/></p>';
}
}
我真的需要知道我怎么可以排序,因为我将使用if和else很多这种说法。任何指导将非常感谢!
预先感谢您!
表达式$ proposalstatus!= 1 || 5'并没有做到你的意思。你必须完全表达这两个条件,所以它看起来像'if($ proposalstatus!= 1 && $ proposalstatus!= 5)',这意味着将其改为'&&'使逻辑工作。 –
'vardump($ proposalstatus);',你应该使用。 – Hackerman
print_r($ proposalstatus)也很有用 –