I'm a beginner and trying to get a handle on php. I have been getting a syntax error that I can't seem to solve. I'll show you the code below and some of the fixes I've tried. If anyone has another idea that would be wonderful. Thank you:)
$subject_set = mysql_query("SELECT * FROM subjects", $connection);
if(!$subject_set){
die("Database query failed: " . mysql_error());
}
while($subject = mysql_fetch_array($subject_set)) {
echo "<li> {$subject['menu_name']} </li>";
}
$page_set = mysql_query("SELECT * FROM pages WHERE id_subjects = {$subject["id"]}", $connection);
if(!$page_set){
die("Database query failed: " . mysql_error());
}
echo "<ul class='pages'>";
while($page = mysql_fetch_array($page_set)) {
echo "<li> {$page['menu_name']} </li>";
}
echo "</ul>";
I get: Database query failed: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near " at line 1
I know the problem is at {$subject["id"]} because I got content back and no error when I put "WHERE id_subjects = 1". I've tried:
{$subject['id']}
{$subject[\"id\"]}
But have gotten the same error...
try
$page_set = mysql_query("SELECT * FROM pages WHERE id_subjects = '".$subject["id"]."'", $connection);
if(!$page_set){
die("Database query failed: " . mysql_error());
}
BTW. you should really move away from mysql_* functions. They are being deprecated, move to PDO or mysqli_*, which are a lot safer as well (you are now vulnerable to sql injection)
If you read back to your post, you can clearly see what's going wrong here.
"SELECT * FROM pages WHERE id_subjects = {$subject["id"]}"
As you can see "id" is not connected to the rest of the rest. That is because with the " you close the string.
To fix this simply use
"SELECT * FROM pages WHERE id_subjects = " . $subject["id"]
Or if you really want to put the variable within the string you can use a single quoted string for the key:
"SELECT * FROM pages WHERE id_subjects = {$subject['id']}"
Personally I am a fan of the first solution. But that is just my opinion.
Well when the while loop finishes looping through, it will have exhausted all the results. $subject['id'] won't have any information simply because $subject no longer has any more entries.
I'm guessing you want to list all the subjects first, then all the pages underneath each subject.
Using mySQL isn't going to be pretty but here's what you want to do. (As Bono said use PDO or mysqli, but here's a solution in psuedocode that will work with mySQL).
loop through first query
print subject name
select pages using subject id
loop through pages under that subject id
print page names
You don't need any quotes when inside a quoted string, just use
"SELECT * FROM pages WHERE id_subjects = {$subject[id]}"
Related
I am following the last part of the following video tutorial "How to create a database website with PHP and mySQL 07 - Add in input form" :
https://www.youtube.com/watch?v=MGIG00d1Xzc&list=PLhPyEFL5u-i0zEaDF0IPLYvm8zOKnz70r&index=7
At the end here is my code, for the inserting portion to the database for the new_jokes.php script (everything up to this point of the series I have gotten to work fine so far)
Basically I am getting the seemingly classic "INSERT INTO" not working although all my syntax looks correct. Am I missing something obvious here? I get no errors, just the row isn't added.
<?php
include "db_connect.php";
$new_joke_question = $_GET["newjoke"];
$new_joke_answer = $_GET["newanswer"];
// Search the database for the word chicken
echo "<h2>Trying to add a new joke and answer: $new_joke_question
$new_joke_answer </h2>";
$sql = "INSERT INTO Jokes_table (JokeID, Joke_question, Joke_answer) VALUES
(NULL, '$new_joke_question', '$new_joke_answer' )";
$result = $mysqli->query($sql);
include "search_all_jokes.php";
?>
Return to the main page
Here is the db_connect.php code as requested:
<?php
// four variables to connect the database
$host = "localhost";
$username = "root";
$user_pass = "usbw";
$database = "test";
// create a database connection instance
$mysqli = new mysqli($host, $username, $user_pass, $database);
?>
Here is search_all_jokes.php (which has minor error checking):
// if there are any values in the table, select them one at a time
if ($mysqli->connect_errno) {
echo "Connection to MySQL failed: (" . $mysqli->connect_errno . ") " .
$mysqli->connect_error;
}
echo $mysqli->host_info . "<br>";
$sql = "SELECT JokeID, Joke_question, Joke_answer FROM Jokes_table";
$result = $mysqli->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "JokeID: " . $row["JokeID"]. " - Joke_question: " .
$row["Joke_question"]. " " . $row["Joke_answer"]. "<br>";
}
} else {
echo "0 results";
}
?>
Also here is the table structure screenshot viewed in myPHPAdmin:
I added error capturing into new_jokes.php inspired by this Stack Overflow post:
INSERT INTO SYNTAX ERROR
And get the following error:
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 't jump.' )' at line 1localhost via TCP/IP
Thank you everyone for helping out with this! Syntax can really throw a wrench in everything. I also will read up on prepared statements since that also could have prevented the issue. The ultimate help to this I found the solution to by adding the function referenced here for MySQLi real_escape_string to clean the single quote I had within the answer I was submitting to my joke table:
(Can a kangaroo jump higher than the empire state building? Of course, the empire state building can't jump.)
As shown in the documentation #miken32 linked as a comment here it is says: "But if $val1 or $val2 contains single quotes, that will make your SQL be wrong. So you need to escape it before it is used in sql; that is what mysql_real_escape_string is for. (Although a prepared statement is better.)"
But now the code for this part 7 of the tutorial on you tube I found works and adds it into a row on the database table, then displaying the full new table on the next webpage. I spent a good while shooting in the dark on while the answer ended up being fairly simple. Again special thanks to #miken32 for pointing me the right direction.
Here is my completed code that ended up working to at least achieve the goal of the tutorial:
<?php
include "db_connect.php";
$new_joke_question = $_GET["newjoke"];
$new_joke_answer = $_GET["newanswer"];
$new_joke_question = $mysqli->real_escape_string($new_joke_question);
$new_joke_answer = $mysqli->real_escape_string($new_joke_answer);
// Search the database for the word chicken
echo "<h2>Trying to add a new joke and answer: $new_joke_question $new_joke_answer
</h2>";
if ($mysqli->connect_errno) {
echo "Connection to MySQL failed: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
echo $mysqli->host_info . "<br>";
$sql = "INSERT INTO Jokes_table (JokeID, Joke_question, Joke_answer) VALUES (' ',
'$new_joke_question', '$new_joke_answer' )";
$result = $mysqli->query($sql);
if ($mysqli->query($sql) === TRUE) {
echo 'users entry saved successfully';
}
else {
echo 'Error: '. $mysqli->error .'<br>';
}
include "search_all_jokes.php";
?>
Return to the main page
Alright. I have searched and searched for an answer, but I just could not find it.
I am writing a simple php script that takes the url information and runs it through a MySQL query to see if a result comes up. I try to echo the variable holding the query out, but nothing shows up. I know there must be a result because if I enter the query manually in MySQL it displays my desired result.
$result = mysqli_query("SELECT * FROM pages WHERE pageq = '" . $_GET['page'] . "'" );
$data = mysqli_fetch_assoc($result);
echo ("You have just entered in " . $data['id'] . "!!! YAY");
I have tried to echo out both the $result and $data. But there is nothing displayed. I am so new to programming, and this is my first StackOverflow post, so forgive me if I am making huge errors.
Actually mysqli_query() requires two parameters... check the following sample example ..
<?php
$conn = mysqli_connect('localhost','root','','your_test_db');
$_GET['page'] = 1;
$result = mysqli_query($conn,"SELECT * FROM your_table WHERE id = '" . $_GET['page'] . "'");
$data = mysqli_fetch_assoc($result);
echo ("You have just entered in " . $data['id'] . "!!! YAY");
?>
As you have stated you are just in a learning phase, it is okay to code these sort of queries just to learn yourself but do not code these kind of queries as these queries are vulnerable so i would suggest you to use prepare queries or PDO...
Also never use SELECT * in your queries, this is a bad practice, only deal with the fields which you requires in return.
Also, you can always check whether your database is connected or not. So that you have a better idea.
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
you have not mentioned whether you are following OOP structure or not .. so i would suggest you to check error_reporting() and connect database on the same page to check the things around ..
Also you can check whether you without WHERE condition for now "SELECT * FROM your_table just to make sure whether you are getting atleast all the records or not.
The problem is that you're not setting up the connection in the query. mysqli_query() requires two parameters.
Make the connection first:
$conn = mysqli_connect("localhost", "user", "password", "dbname");
Now execute the query:
$result = mysqli_query($conn,"SELECT * FROM pages WHERE pageq = '" . $_GET['page'] . "'" );
NOTE: Your code is heavily vulnerable to MySQL injections. Use MySQLi or PDO Prepared statements.
Also, you should use mysqli_errno() to find out your query bugs.
Edit:
Also do this:
while($row=mysqli_fetch_assoc($result)){
//do the result output.
}
I have code here that is supposed to print a html table from my mysql database. When I open the page in my web browser, it is a blank page.
<html>
<body>
<?php
$connection = mysql_connect('localhost', 'admin', 'may122000');
mysql_select_db('contacts');
$query = "SELECT * FROM users";
$result = mysql_query($query);
echo "<table>"; // start a table tag in the HTML
while($row = mysql_fetch_array($result)){
echo "<tr><td>" . $row['first_name'] . "</td><td>" . $row['last_name'] . "</td></tr>"; //$row['phone'] the index here is a field name
}
echo "</table>";
mysql_close();
?>
</body>
</html>
Remove password
Enable error output
When you use mysql_fetch_array you will get the resulting array with numeric indices.
mysql_fetch_assoc will give you an associative array, like you want.
Note: mysql_* is deprecated.
while($row = mysql_fetch_assoc($result)){
echo "<tr><td>" . $row['first_name'] . "</td><td>" . $row['last_name'] . "</td></tr>"; //$row['phone'] the index here is a field name
}
If you still want to use mysql_fetch_array you'll have to pass a second parameter:
while($row = mysql_fetch_array($result, MYSQL_ASSOC)){
First of all user mysqli or PDO and mysqli_fetch_assoc() so you have only associative array. Blank page is probably result of a hidden error, that's stored in your error.log on your server - take a look at it and get back to us.
I prefer using PDO or mysqli but anyway , Are u sure Your connection is established ? to check this and check other connections and query :
if (!connection)
die(mysql_error());
try this and feedback me
Improvements - some of which already mentioned in other post but all put together in one form:
<?php
$connection = mysqli_connect('localhost', 'admin', '****', 'contacts');
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$query = "SELECT first_name, last_name, phone FROM users";
$result = mysqli_query($connection, $query) or die(mysqli_error($connection));
echo "<table>"; // start a table tag in the HTML
while($row = mysqli_fetch_array($result)){
echo "<tr><td>" . $row['first_name'] . "</td><td>" . $row['last_name'] . "</td></tr>"; //$row['phone'] the index here is a field name
}
echo "</table>";
mysqli_close($connection);
?>
So, first off the MySQL_* has been upgraded to Mysqli, with some minor reformatting,
The select * has been replaced with selecting only the needed columns.
The closing statement has been correctly set.
Firstly if your connection fails an error catch will output this to the screen. Remove this upon product launch or public launch of the page.
A (Rather rudimentary) error catch has been put in that if the SQL Query is bad that an error is outputted. Again, this should be removed in production but will help you with finding SQL errors.
If No SQL errors return the you have either an empty table in your database, or some sort of PHP error but from the code sample given the most likely error is that your PHP doesn't run MySQL and would only run PDO or MySQLi.
You also said "when I open the page in my browser it is a blank page", if the Source of the page is blank - as in it DOES NOT show
<html>
etc, then this is a sign the PHP execution failed and you have bad PHP, as detailed in your error log file.
The most likely cause of this from the code sample given is, as stated already, your PHP version does not support MySQL.
If your
<table>
Tag appears in your HTML source code then this is a sign that the While clause is not running which means your Datbase table is empty and there is no data to output.
Hope this helps. But first point of call is to upgrade to MySQLi :)
I am brand new to php/mysql, so please excuse my level of knowledge here, and feel free to direct me in a better direction, if what I am doing is out of date.
I am pulling in information from a database to fill in a landing page. The layout starts with an image on the left and a headline to the right. Here, I am using the query to retrieve a page headline text:
<?php
$result = mysql_query("SELECT banner_headline FROM low_engagement WHERE thread_segment = 'a3'", $connection);
if(!$result) {
die("Database query failed: " . mysql_error());
}
while ($row = mysql_fetch_array($result)) {
echo $row["banner_headline"];
}
?>
This works great, but now I want to duplicate that headline text inside the img alt tag. What is the best way to duplicate this queries information inside the alt tag? Is there any abbreviated code I can use for this, or is it better to just copy this code inside the alt tag and run it twice?
Thanks for any insight!
You are, as the comment says, using deprecated functions, but to answer your question, you should declare a variable to hold the value once your retrieve it from the database so that you can use it whenever your want.
<?php
$result = mysql_query("SELECT banner_headline FROM low_engagement WHERE thread_segment = 'a3'", $connection);
if(!$result) {
die("Database query failed: " . mysql_error());
}
$bannerHeadline = "";
while ($row = mysql_fetch_array($result)) {
$bannerHeadline = $row["banner_headline"];
}
echo $bannerHeadline; //use this wherever you want
?>
It is hard to help without knowing more. You are pumping the results into an array, are you expecting to only return one result or many banner_headline results? If you will only ever get one result then all you need to do is something like this:
PHP:
$result = mysql_query("
SELECT `banner_headline`
FROM `low_engagement`
WHERE `thread_segment` = 'a3'", $connection) or die(mysql_error());
// This will get the zero index, meaning first result only
$alt = mysql_result($result,0,"banner_headline");
HTML:
<html>
<body>
<!--- Rest of code -->
<img src="" alt="<?php echo $alt ?>">
On a side note, you should stop using mysql-* functions, they are deprecated.
You should look into PDO or mysqli
I'm sorry, probably somewhere there will be the answer to my question, but it's hours I'm looking for trying to resolve this problem:
Here is the code:
<?php
$con = mysql_connect("****","****","***");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("******", $con);
$query = "SELECT id FROM fq_questions";
$rowcnt = mysql_num_rows(mysql_query($query));
echo "Tot scenes: ".$rowcnt;
$id = rand (1,$rowcnt);
echo "<br>Id rand: ".$id."<br>";
flush();
$newquery = "SELECT question FROM fq_questions WHERE id=".$id;
$result = mysql_query($newquery);
if (!$result) {
$message = 'Invalid query: ' . mysql_error() . "\n";
$message .= 'Whole query: ' . $newquery;
die($message);
}
$row = mysql_fetch_assoc($result);
echo $row['question'];
mysql_close($con);
?>
The problem is that there is no output. I've tried everything, seems to be a problem into the query, but there is a result, it's not false, but even if it exists, nothing is outputted.
The code works till
echo "<br>Id rand: ".$id."<br>";
then it shows nothing.
It's a dummy problem, i'm getting crazy just because of it.
Uh, was forgetting... The website where I've got the problem: http://www.freelabs.it/filmquiz/game.php
Be careful, desc is a SQL keyword ! Your query may not compile because of that.
"desc" is a MySQL reserved word, you should just change that column name in your DB. Anyway, your method is not random at all, and it will fail as soon as you have a "hole" in your ids (when you delete one member).
Take a look at MySQL "ORDER BY RAND()"
$data = mysql_query("SELECT description FROM fq_questions ORDER BY RAND() LIMIT 1");
You used mysql_fetch_row() which returns a numerical array. You then try to access the array slot named 'desc'.
It doesn't exist. (My guess is that that produces a supressed error, preventing any output from showing up, or a supressed warning, preventing any output after that line from showing up.)
Try changing mysql_fetch_row() to mysql_fetch_assoc() (still DEPRECATED!) and that should be solved.
Sources: http://php.net/manual/en/function.mysql-fetch-row.php
And: http://php.net/manual/en/function.mysql-fetch-assoc.php