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誰かが私の$cat_id変数がどのように認識されるのか教えてください。第二部私のコードは？それは、FIRST PARTで正常に動作します - 私は$cat_id値がでmysqlデータベースに挿入されている意味：PHP。私の変数はスコープ全体で認識されないようです。何故なの？
$insert_review_command = "INSERT INTO review VALUES(NULL,'$cat_id','{$category}','$user_id', '{$name}','{$phonenumber}','{$address}', '{$comment}')";
しかし、何が第二部に挿入されていません。私は$cat_idが認識されているとは思わない。しかし、なぜそうではありませんか？$cat_idは私のコード全体で認識されませんか？ If statement内で定義されている場合は、それはIf statement内でのみ認識されますか？助けてくれてありがとう。
<?php require('dbConnect.php');
if (isset($_POST['create'])) {
$category = ($_POST['category']);
$name = ($_POST['name']);
$phonenumber = ($_POST['phonenumber']);
$address = ($_POST['address']);
$comment = ($_POST['comment']);
//check if the category being entered is already there
$select_from_cat_table = "SELECT * FROM category WHERE cat_name = '$_POST[category]'";
$result=mysqli_query($con,$select_from_cat_table);
$num_rows = mysqli_num_rows($result);
// get the matching cat_id
$row = mysqli_fetch_assoc($result);
$cat_id = $row["cat_id"];
// ****FIRST PART**** $CAT_ID IS INSERTED INTO THE DB
//if the category name already exists in the category table, then don't add it in again
if($num_rows >= 1) {
echo "This Already Exists<br/>";
//but do add it to the review table
//for the cat_id, we want to get the cat_id of the category name that already exists, that has
//just been posted. This is $cat_id. $user_id is the user id of the person posting
$insert_review_command = "INSERT INTO review VALUES(NULL,'$cat_id','{$category}','$user_id', '{$name}','{$phonenumber}','{$address}', '{$comment}')";
$insert_into_review_table = mysqli_query($con,$insert_review_command);
}
// ****SECOND PART**** $CAT_ID IS NOT INSERTED INTO THE DB
else if ($num_rows < 1)
{
//if it's not in there, then add the category in the category table.
$insert_category_command = "INSERT INTO category VALUES(NULL, '{$category}', '$user_id')";
$insert_into_category_table = mysqli_query($con,$insert_category_command);
//****WHY IS CAT_ID NOT WORKING HERE????******
//and add it to the review table
//for the cat_id, we want to get the cat_id of the category name that already exists, that has
//just been posted. This is $cat_id. $user_id is the user id of the person posting
$insert_review_command = "INSERT INTO review VALUES(NULL,'$cat_id','{$category}','$user_id', '{$name}','{$phonenumber}','{$address}', '{$comment}')";
$insert_into_review_table = mysqli_query($con,$insert_review_command);
echo "Yes, it's been added correctly";
echo $cat_id;
}
$con->close();
header('Location:volleyLogin.php');
}
?>
<!doctype html>
<html>
<body>
<h2>Create new Contact</h2>
<form method="post" action="" name="frmAdd">
<p><input type="text" name = "category" id = "category" placeholder = "category"></p>
<p><input type="text" name = "name" id = "name" placeholder = "name"></p>
<p><input type="text" name = "phonenumber" id = "phonenumber" placeholder = "phone number"></p>
<p><input type="text" name = "address" id = "address" placeholder = "address"></p>
<p><input type="text" name = "comment" id = "comment" placeholder = "comment"></p>
<h2>Visible to :</h2>
<input type="radio" name="allmycontacts" value="All my Contacts">All my Contacts
<input type="radio" name="selectwho" value="Select Who">Select Who
<input type="radio" name="public" value="Public">Public
<input type="radio" name="justme" value="Just me">Just me
<p><input type="submit" name = "create" id = "create" value = "Create new Contact"></p>
<a href="exit.php">Exit</a>
</form>
</body>
</html>
'（$ num_rows <1）'が真を返すので、 '$ row = mysqli_fetch_assoc（$ result）;'は何も返しませんでした。そして、データベースから '$ cat_id'を取得するので、フェッチする行がない場合でも、値を持つことはできません。 – Qirel
'$ insert_into_category_table = ...'、*の前、* '$ insert_review_command = ...'の前に、 '$ cat_id = mysqli_insert_id（$ con）;'を使って、ちょうど挿入されたカテゴリIDを取得します。 [mysqli_insert_id（） ']（http://php.net/manual/en/mysqli.insert-id.php） – Sean
また、" SELECT * FROMカテゴリWHERE cat_name = '$ _POST [category]' "; 「SELECT * FROM category WHERE cat_name = '$ category'」とすることができます。 –