This question already has an answer here:
Syntax error due to using a reserved word as a table or column name in MySQL
(1 answer)
Closed 7 years ago.
I am extremely new to PHP and I'm frustrated by the error of this simple task. I want to import a table from an SQL database and show it in a HTML table. But I keep getting errors when trying to fetch the table column names.
The connection with the database is made though, I tested that.
The error is:
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 'Table' at line 1.
I found this example on w3schools which I edited by other examples on php.net
If anybody can help me with this, I'd appreciate it.
<?php
error_reporting(-1);
$con = mysqli_connect('localhost', 'user', 'pass');
mysqli_select_db($con, 'database') or die("Could not found" . mysqli_error($con));
$query = ("select * from Table");
$result = mysqli_query($con, $query) or die ( mysqli_error ($con) );
//Print table
echo "<table>";
echo "<tr>";
if($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
//Print headers
foreach($row as $key => $value ){
echo "<td>" . $key . "</td>";
}
echo "</tr>";
}
$result = mysqli_query($con, $query) or die (mysqli_error());
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
echo "<tr>";
while( list($key, $value) = each($row)){
//Print value
echo "<td>" . $value . "</td>";
}
echo "<td>" . $value . "<i class='fa fa-caret-up'></i><i class='fa fa-caret-down'></i></td>";
echo "</tr>";
}
echo "</table>";
?>
Table is a reserved keyword and you cant use it like this. If you want to fetch some data from a table named users or some like this then the query should be -
select * from users
Related
This question already has an answer here:
Syntax error due to using a reserved word as a table or column name in MySQL
(1 answer)
Closed 8 years ago.
This code is supposed to show the number of rows in the column "id". Why isn’t is working, when I go to the page it shows nothing except my HTML stuff?
<?php
$con = mysql_connect("quollcraft.net", "quollcr1_forum", "password");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("quollcr1_hub", $con);
$result = mysql_query("SELECT id FROM table ORDER BY id DESC LIMIT 1 ");
echo "<table>";
while ($row = mysql_fetch_array($result)) {
echo "<td>";
echo "<center>";
echo '<p>' . $row ['id'] . ' total users<p>';
echo "</td>";
}
echo "</table>";
mysql_close($con);
?>
Your table name you are using are just table. I guess it should be changed to the real table name from which you want your data
table is one of the reserved word(s) in MySQL. You need to wrap them using backticks.
Like this..
$result = mysql_query("SELECT id FROM `table` ORDER BY id DESC LIMIT 1 ");
^ ^ //<---- Like that
This (mysql_*) extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, Prepared Statements of MySQLi or PDO_MySQL extension should be used to ward off SQL Injection attacks !
Others have decent MySQL tips, but what happens when you view the source code? Because the HTML seems to be broken at best. This is the cleaned up version based on what I am seeing.:
echo "<table>";
echo "<tr>";
while ($row = mysql_fetch_array($result)) {
echo "<td>";
echo "<center>";
echo '<p>' . $row ['id'] . ' total users<p>';
echo "</center>";
echo "</td>";
}
echo "</tr>";
echo "</table>";
You were missing the table row tags <tr> & </tr> as well as the closing </center> tag.
Looks like you are trying to retrieve the total count of ID(users) but you missed out the keyword count. You can modify either the query or you can get the count of $result. Hope this helps.
I am still a beginner with php and MySQL. I am having trouble getting rows from my database to display in an html select drop down box. I have researched it and it seems like my code should be good. The campaigns table as a row titled name. This is the row I am wanting to echo into the drop down. The drop down shows, however there is no content in it. Not sure what I am missing here...
Here is the code
<?php
$con=mysqli_connect("localhost","username","password","db_name");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$query = mysqli_query($con,"SELECT * FROM campaigns");
echo '<select name="campaignChange">';
while ($row = mysql_fetch_array($query)) {
echo "<option value='" . $row['name'] ."'>" . $row['name'] ."</option>";
}
echo '</select>';
?>
You are mixing mysql and mysqli syntax.
You should change:
$query = mysql_query($con,"SELECT * FROM campaigns");
to:
$query = mysqli_query($con,"SELECT * FROM campaigns");
and:
while ($row = mysql_fetch_array($query)) {
to:
while ($row = mysqli_fetch_array($query)) {
By the way, you should add error handling. If you add this to the top:
mysqli_report(MYSQLI_REPORT_ALL);
mysqli will throw exceptions so you will always know what goes wrong exactly. As long as you use mysqli functions of course...
I have the code bellow, it's ok but I want to be able to use for example the 4th value extracted from the database, use it alone, not put all of them in a list, I want to be able to use the values from database individually. How do I echo them?
Edit: I was thinking to simplify things, to be able to add the values from database into one array and then extract the value I need from the array (for example the 4th - ordered by "order_id"). But how?
Right now I can only create a list with all the values one after the other..
(Sorry, I am new to this). Thank you for all your help..
<?php
include '../../h.inc.php';
$con = mysql_connect($db['host'],$db['user'],$db['passwd']);
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("database", $con);
$result = mysql_query("SELECT * FROM options WHERE Name LIKE 'x_swift%' ORDER BY order_id");
echo "<table border='1'>
<tr>
<th>Values</th>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
// echo "<td>" . $row['Name'] . "</td>";
echo "<td>" . $row['VALUE'] . "</td>";
echo "</tr>";
$array = array(mysql_fetch_array($strict));
}
echo "</table>";
mysql_close($con);
?>
To select the value in the value column of the row where order_id is 4, use this SQL:
$query = 'select value from options where order_id = 4';
Then you can access this result in many ways. One is to get the entire result row (which in this case is just one cell) as an associative array:
if ($result = mysql_query($query)) {
$row = mysql_fetch_assoc($result);
echo 'value = ' . $row['value'];
}
You can also get the value directly:
if ($result = mysql_query($query)) {
echo 'value = ' . mysql_result($result, 'value');
}
It would just be a query like...
$result = mysql_query("SELECT * FROM options WHERE ID = 3");
mysql_fetch_row($result);
Unless Im misunderstanding you....let me know
But you really should use PDO, instead of deprecated mysql_* functions
http://php.net/manual/en/book.pdo.php
This question already has answers here:
How can I use PDO to fetch a results array in PHP?
(2 answers)
Closed 2 years ago.
I have a php script that selects data via mysql_, however recently I have been reading that PDO is the way to go and that mysql_ is becoming depreciated. Now I am converting that script to PDO.
My question is though, I am not using $_POST to select. I just want to select the entire table with all of its data so I enter this query :
$query = $dbh->prepare("SELECT * FROM students");
$query->execute();
$result = $query->fetchall(); // or you can just $result = $query as hakre proposed!
so then like I did with my old depreciated mysql_ version of the script I used the echo to echo a table with the data in it.
echo
"<table border='2'>
<tr>
<th>ID</th>
<th>A Number</th>
<th>First Name</th>
<th>Last Name</th>
<th>Why</th>
<th>Comments</th>
<th>Signintime</th>
</tr>"
;
foreach($result as $row)
{
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" .$row['anum'] . " </td>";
echo "<td>" . $row['first'] . "</td>";
echo "<td>" . $row['last'] . "</td>";
echo "<td>" . $row['why'] . "</td>";
echo "<td>" . $row['comments'] . "</td>";
echo "<td>" . $row['signintime'] . "</td>";
echo "<td> <input type=\"button\" value=\"Start Session\"onClick=\accept.php?id=" . $row['id'] . "&start=true></td>";
}
echo "</tr>";
echo "</table>";
now using this, I can not get a single output to my table.
My question is am I missing something from my select statements? Or am I not fetching any rows? Also I the connection settings set in another script called connect.php that is required by init.php (at the top of all of my pages)
Edit : 1
Edited the code so it now works, also adding a picture to show others how it should look! Hopefully some one can put this to some sort of use!
You are doing too much actually:
$query = $dbh->prepare("SELECT * FROM students");
$query->execute();
$result = $dbh->query($query);
The problematic line is:
$result = $dbh->query($query);
Check with http://php.net/pdo.query, the parameter is a string, actually the SQL string you already use above, not the result value of a PDO::prepare() call.
For your simple query you can just do:
$result = $dbh->query("SELECT * FROM students");
Or if you like to prepare:
$query = $dbh->prepare("SELECT * FROM students");
$query->execute();
$result = $query;
The later is some boilerplate if you want to insert variables into the query, that's why you prepare it.
The next problem is with the foreach line:
foreach($result as $row);
You are terminating the loop immediately because of the semicolon ; at the end. Remove that semicolon so that the following angle-bracketed code-block becomes the body of the foreach-loop.
Your code is wrong:
$query = $dbh->prepare("SELECT * FROM students");
$query->execute();
$result = $dbh->query($query);
After executing a prepared statement, you can just call fetchAll() on it:
$query = $dbh->prepare("SELECT * FROM students");
$query->execute();
$result = $query->fetchAll();
The rest of your code will work fine once you remove the semicolon after the foreach.
Basicaly having issues setting up a webpage which will taken in a student key entered by the user. This will then parse the student key to another file which will run it against a mysql backend to see what records this student already has. But can not get it working for the life of me please help I'm still a newb at this.
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("support_log", $con);
$result= mysql_query("SELECT student.first_name, student.surname, student.year_group, student.STKEY, student_log.issue
FROM `student` JOIN `student_log`
WHERE student.STKEY like '$_POST[stkey]'");
$result2 = mysql_query($result) or die("Error: " . mysql_error());
if(mysql_num_rows($result2) == 0){
echo("no records found");
} ELSE {
echo "<table border='1'>
<tr>
<th>First name</th>
<th>Surname</th>
<th>Year Group</th>
<th>Student Key</th>
<th>Issue</th>
</tr>";
while($row = mysql_fetch_array($result2))
{
echo "<tr>";
echo "<td>" . $row['First_Name'] . "</td>";
echo "<td>" . $row['surname'] . "</td>";
echo "<td>" . $row['year_group'] . "</td>";
echo "<td>" . $row['stkey'] . "</td>";
echo "<td>" . $row['issue'] . "</td>";
echo "</tr>";
}
echo "</table>";
}
mysql_close($con);
?>
After changing my where statement to:
WHERE student.STKEY like '$_POST[stkey]'");
I am no longer reciving errors from PHP but now recieving the error Query was empty which is part of my code to detect if there is no results. Though I have tested that query in phpmyadmin and it spits out results. From looking at the code does anyone have any solutions? I have also checked the parse by running an echo on the post command to ensure the data being entered was correct.
Edit: Got rid of the whole result2 check now throwing a:
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\stkey_submit.php on line 24
Try $_POST['stkey'] instead of $_POST[stkey]
EDIT : if you use it in a query, it would be preferable to do :
$stkey = mysql_real_escape_string($_POST['stkey']);
$sql = "SELECT ....... like '$stkey'";
mysql_query($sql);
$result= mysql_query("SELECT student.first_name, student.surname, student.year_group, student.STKEY, student_log.issue
FROM `student` JOIN `student_log`
WHERE student.STKEY like " . $_POST["stkey"]);
How about storing the value of stkey on a variable before including it on the query?
$stkey = $_POST['stkey'];
$result= mysql_query("SELECT student.first_name, student.surname,
student.year_group, student.STKEY, student_log.issue
FROM `student` JOIN `student_log`
WHERE student.STKEY LIKE '%$stkey%'");
You might also want to use MySqli or PDO instead of the MySql database API. Take a look at this post from Nettuts: http://net.tutsplus.com/tutorials/php/php-database-access-are-you-doing-it-correctly/