Parse error: syntax error, unexpected '$run' (T_VARIABLE) in C:\xampp\htdocs\test\admin\includes\insert_post.php on line 12

Question

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I am new OOP PHP learner.I just wanted to create a table and insert my posts.Then I've faced Parse error above.

What I have tried:

PHP

<pre><?php
include "db.php";
include "config.php";

$db = new database();

if (isset($_POST['submit'])){
	$title = $_POST['title'];
	$content = $_POST['content'];
$query = "INSERT INTO posts(title,content) VALUES('$title',$content')"

$run=$db->insert($query);

}

?>



<!DOCTYPE html>
<html>
<head>
	<title></title>
</head>
<body>

</body>
</html>

<form action="insert_post.php" method="post" enctype="multipart/form-data">
	<div class="form-group">
	<center><table width="800" align="center" border="2">
		<tr bgcolor="orange">
		<td colspan="6"><h1 style="text-align:center">Post EDIT</h1></td>
		</tr>
		<tr>
			<td align="right" bgcolor="orange">Post title:</td>
			<td><input type="text" name="title" size="60"></td>
		</tr>
		<tr>
			<td align="right" bgcolor="orange">Post Content:</td>
			<td><textarea name="Content" rows="15" cols="40"></textarea></td>
		</tr>
     <tr>

	<td colspan="6" align="center" bgcolor="orange"><input type="Submit" name="submit" value="Publish Now"  /></td>
</tr>	






	</table>






</form>

Posted 17-Oct-17 3:40am

ALEX8998

Updated 29-Apr-22 0:43am

Add a Solution

3 solutions

Solution 6

In addition to solution 1, I suspect

PHP

$query = "INSERT INTO posts(title,content) VALUES('$title','$content')";
// a single quote was missing                         here ^

Never build an SQL query by concatenating strings. Sooner or later, you will do it with user inputs, and this opens door to a vulnerability named "SQL injection", it is dangerous for your database and error prone.
A single quote in a name and your program crash. If a user input a name like "Brian O'Conner" can crash your app, it is an SQL injection vulnerability, and the crash is the least of the problems, a malicious user input and it is promoted to SQL commands with all credentials.
SQL injection - Wikipedia[^]
SQL Injection[^]
SQL Injection Attacks by Example[^]
PHP: SQL Injection - Manual[^]
How can I explain SQL injection without technical jargon? - Information Security Stack Exchange[^]

Posted 29-Apr-22 0:43am

Patrice T

v2

Comments

CHill60 29-Apr-22 10:40am

Did you notice the date of the OP?

Patrice T 29-Apr-22 13:08pm

No Missed it. Thank you

Solution 10

mysqli_select_db($con,"murid");
$hasil = mysqli_query($con,"SELECT * FROM MURID
WHERE NOKP= '".$nokp."'") ;
$row = mysqli_fetch_array($hasil);
$nama = htmlspecialchars($row['NAMA'],ENT_QUOTES);
$notel = $row['NOTEL'];
$emel =$row['EMEL'];
$idmurid =$row['IDMUIRD'];

Posted 1-Jul-22 4:57am

Member 15693792

Updated 1-Jul-22 5:24am

Comments

Dave Kreskowiak 1-Jul-22 11:49am

Whatever this is it's getting removed as a non-answer to a five year old question.

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Jochen Arndt · Accepted Answer · 2017-10-17T04:10:00

Solution 1

There is no semicolon at the end of the previous line:

PHP

$query = "INSERT INTO posts(title,content) VALUES('$title',$content')";
//                                       This is missing in your code ^

See also PHP: Instruction separation - Manual[^]

Posted 17-Oct-17 4:10am

Jochen Arndt