Good day. I'm new, trying to create a form where you can insert data into the database and display it on the page. I have 2 tables in the database - film and category. The code is as follows:

<?php // login page with php code require_once 'log.php'; if (isset($_POST['delete'])) { $id = get_post($conn, 'id'); $query = "DELETE FROM film WHERE id='$id'"; $result = $conn->query($query); if (!$result) echo "DELETE failed: $query<br>" . $conn->error . "<br><br>"; } if (isset($_POST['title']) && isset($_POST['director']) && isset($_POST['year']) && isset($_POST['type'])) { $title = mysql_real_escape_string($_POST['title']); $director = mysql_real_escape_string($_POST['director']); $year = mysql_real_escape_string($_POST['year']); $type = mysql_real_escape_string($_POST['type']); $query1 = "INSERT INTO film (title, director, year) VALUES ($title', '$director', '$year')"; $query2 = "INSERT INTO category (type) VALUES ('$type')"; $result1 = $conn->query($query1); $result2 = $conn->query($query2); if (!$result1) echo "INSERT failed: $query1<br>" . $conn->error . "<br><br>"; if (!$result2) echo "INSERT failed: $query2<br>" . $conn->error . "<br><br>"; } echo <<<_END <form action="" method="post"><pre> Title <input type="text" name="title"> Director <input type="text" name="director"> Year <input type="text" name="year"> Type <input type="text" name="type"> <input type="submit" value="ADD RECORD"> </pre></form> _END; echo "<table>"; echo "<tr> <th>Title</th> <th>Director</th> <th>Year</th> <th>Genre</th> <th>Update</th> <th>Delete</th></tr>"; $query = "SELECT title,director,year,type FROM film JOIN category ON category.catID=film.catID"; $result = $conn->query($query); if (!$result) die ("Database access failed: " . $conn->error); $rows = $result->num_rows; for ($j = 0 ; $j < $rows ; ++$j) { $result->data_seek($j); $row = $result->fetch_array(MYSQLI_NUM); echo "<tr>"; echo '<td>' . $row[0] . '</td>'; echo '<td>' . $row[1] . '</td>'; echo '<td>' . $row[2] . '</td>'; echo '<td>' . $row[3] . '</td>'; echo '<td>Edit</td>'; echo '<td><button type="delete" name="delete" value="delete">delete</button></td>'; echo "</tr>"; } echo "</table>"; ?>

Gives an error message

"INSERT failed: INSERT INTO film (title, director, year) VALUES ('h', 'h', '1890')"

What's wrong?

Update

 <?php // login page with php code require_once 'log.php'; if (isset($_POST['delete']) && isset($_POST['id'])) { $id = get_post($conn, 'id'); $query = "DELETE FROM film WHERE id='$id'"; $result = $conn->query($query); if (!$result) echo "DELETE failed: $query<br>" . $conn->error . "<br><br>"; } if (isset($_POST['title']) && isset($_POST['director']) && isset($_POST['year']) && isset($_POST['type'])) { $title = mysql_real_escape_string($_POST['title']); $director = mysql_real_escape_string($_POST['director']); $year = mysql_real_escape_string($_POST['year']); $type = mysql_real_escape_string($_POST['type']); $query = "INSERT INTO film (title, director, year, type) VALUES ('".$title."', '".$director."', '".$year."', '".$type."')"; $result = $conn->query($query); if (!$result) echo "INSERT failed: $query<br>" . $conn->error . "<br><br>"; } echo <<<_END <form action="" method="post"><pre> Title <input type="text" name="title"> Director <input type="text" name="director"> Year <input type="text" name="year"> <input type="radio" name="type" id="drama" value="drama" checked> drama<br> <input type="radio" name="type" id="mystery" value="mystery"> mystery<br> <input type="radio" name="type" id="thriller" value="thriller"> thriller <br> <input type="radio" name="type" id="comedy" value="comedy"> comedy <input type="submit" value="ADD RECORD"> </pre></form> _END; echo "<table>"; echo "<tr> <th>ID</th> <th>Title</th> <th>Director</th> <th>Year</th><th>Category</th><th>Update</th> <th>Delete</th></tr>"; $query = "SELECT id,title,director,year,type FROM film"; $result = $conn->query($query); if (!$result) die ("Database access failed: " . $conn->error); $rows = $result->num_rows; for ($j = 0 ; $j < $rows ; ++$j) { $result->data_seek($j); $row = $result->fetch_array(MYSQLI_NUM); echo '<tr>'; echo '<td>' . $row[0] . '</td>'; echo '<td>' . $row[1] . '</td>'; echo '<td>' . $row[2] . '</td>'; echo '<td>' . $row[3] . '</td>'; echo '<td>' . $row[4] . '</td>'; echo '<td>Edit</td>'; echo '<td><form action="" method="post"> <input type="hidden" name="delete" value="yes"> <input type="hidden" name="id" value="$row[id]"> <input type="submit" value="delete"></form></td>'; echo '</tr>'; } echo "</table>"; $result->close(); $conn->close(); function get_post($conn, $var) { return $conn->real_escape_string($_POST[$var]); } ?>

This is a reworked version. I removed the second table from the database, filling out the form works, but deleting does not. Where is the mistake?

Closed due to the fact that off-topic participants cheops , Ipatiev , zRrr , user194374, aleksandr barakin 21 Jun '16 at 6:20 .

It seems that this question does not correspond to the subject of the site. Those who voted to close it indicated the following reasons:

  • "The question is caused by a problem that is no longer reproduced or typed . Although similar questions may be relevant on this site, solving this question is unlikely to help future visitors. You can usually avoid similar questions by writing and researching a minimum program to reproduce the problem before publishing the question. " - cheops, aleksandr barakin
  • “Questions asking for help with debugging (“ why does this code not work? ”) Should include the desired behavior, a specific problem or error, and a minimum code for playing it right in the question . Questions without an explicit description of the problem are useless for other visitors. See How to create minimal, self-sufficient and reproducible example . " - Ipatiev, zRrr, Community Spirit
If the question can be reformulated according to the rules set out in the certificate , edit it .

  • one
    A ready-made working example, without all the horror that they advised below, phpfaq.ru/safemysql/crud - Ipatyev
  • @Ipatiev Please try to publish detailed answers containing a specific example of the minimum solution, supplementing them with a link to the source. Answers –references (like comments) do not add knowledge to Runet. - Nicolas Chabanovsky ♦
  • @Nic I have never seen such a rule regarding comments. Nobody can reproach me about the developed answers for not writing them. If I see that the question is worth it. For the offtopic question, like, "I’ve been rolling 5 screens of code here, find me a mistake in it," I value my time too much, since no answer to it will "add knowledge to the Runet." - Ipatiev
  • @ Ipatiev Please vote for closing, in this case. - Nicolas Chabanovsky ♦

1 answer 1

In sql queries you need to correctly enter variables. For example, so.

 $query1 = "INSERT INTO film (title, director, year) VALUES ('".$title."', '".$director."', ".$year.")"; $query2 = "INSERT INTO category (type) VALUES ('".$type."')";

Just in case, check if the data types match (variable and table fields) for title , director , year and type .

PS: For better readability, PHP variables can be inserted into the query like this:

 $query1 = "INSERT INTO film (title, director, year) VALUES ('{$title}', '{$director}', '{$year}')"; $query2 = "INSERT INTO category (type) VALUES ('{$type}')";
  • one
    Yes, the best option ('{$ param}') is tCode
  • 2
    It’s probably useless to ask what’s wrong with “entering the variables” in the original question. - Ipatiev
  • one
    php.net/manual/en/pdostatement.bindparam.php - E_p