I'm trying to use this query
$page_set = mysql_query("SELECT * FROM pages WHERE subject_id =
{$subject["id"]}", $connection);
but i keep getting this error when loading my page .
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
Try it without the complex syntax:
$query = 'SELECT * FROM pages WHERE subject_id = ' . $subject['id'];
$page_set = mysql_query($query, $connection);
Incidentally, I loathe variable parsing in strings, and prefer concatenation.
you're experiencing a quote mismatch. try replacing the double quotes around your array key with single quotes.
$page_set = mysql_query("SELECT * FROM pages WHERE subject_id =
{$subject['id']}", $connection);
$sql = "SELECT * FROM pages WHERE subject_id = '".$subject["id"]."'";
$page_set = mysql_query($sql, $connection);
Make sure you escape the subject_id also.
use single quote
Related
$select = "SELECT name FROM table_name WHERE location ='".$loc."' ";
$findname = mysql_query($select) or die(mysql_error());
I keep getting this error! I have tried everything!!!
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 'WHERE location ='Florida'' at line 1
$loc is determined by the following:
<input type="text" name="loc"> in the HTML
$loc = $_POST['loc']; in the PHP
Try this , use mysql_escape_string or mysql_real_escape_string mysql safe string functions
$select = "SELECT `name` FROM `table_name` WHERE `location` ='".mysql_escape_string($loc)."' ";
$findname = mysql_query($select) or die(mysql_error());
$select = "SELECT name FROM table_name WHERE location ='".mysql_real_escape_string($loc)."' ";
$findname = mysql_query($select) or die(mysql_error());
Just check for any single quotes or double quotes in the location variable. That might be a problem.
Use str_replace(find,replace,string) to replace single and double quotes in the string.
Example, when you can contain a double quote in the $loc variable.
$select = "SELECT name FROM table_name WHERE location ='Flo"rida' ";
The query will end at Flo.
use mysql_real_escape_string after WHERE location =
I fixed it. I was using a variable for the table name and was referencing a null value. Thanks for the help!
My Code looks like below.
$var = 'ID="'. mysql_real_escape_string($data[0]).'" AND SYS="'.mysql_real_escape_string($data[2]). '" AND TITLE="'.mysql_real_escape_string($data[1]).'"';
$sql = 'SELECT * FROM `table_name` WHERE '. $var;
$result = mysql_query($sql);
In the where condition, TITLE when using a single quote(') I am facing the below error even though the mysql_real_escape_string() function is being used.
The error thrown is
Resource id #5You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 's Created', 'Test', 'Test', '0000-00-00 00:00:00', ' at line 25
after your dumping looks like you have problem with apostroph
you may change your quotes like that
$var = "ID='". mysql_real_escape_string($data[0])."' AND SYS='".mysql_real_escape_string($data[2]). "' AND TITLE='".mysql_real_escape_string($data[1])."' ";
$sql = "SELECT * FROM `table_name` WHERE ". $var;
$result = mysql_query($sql);
$finalvar=stripslashes($var);
$sql = 'SELECT * FROM table_name WHERE '. $finalvar;
Try dumping your SQL query in its compete form right before it is sent.
You'll be able to spot the error that way.
I have php script like this
$query = "select * where userid = 'agusza' ";
$result = mysql_query($query) or die(mysql_error());
while($row=mysql_fetch_array($result)) {
echo $result;
}
when I execute, the result like this
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 'where userid = 'agusza'' at line 1
But when I run that sql in sqlserver, it running well
Anybody has solution ?
$query = "select * from table_name where userid = 'agusza' ";
See the corrections I have made. You haven't used the right syntax for SELECT query
You didn't select a table using FROM. Without that, it does not know which table you are selecting data from.
You should also stop using mysql as it is deprecated. Use mysqli or PDO as they are safer.
You are also echoing the wrong variable in your while loop, try this:
while ($row = mysql_fetch_array($result) {
echo $row['column_name'];
}
$query = "select * from table where userid = 'agusza'";
Right now, you're not telling which table SQL should look in.
You should format your query like so:
select * from `TableName` where userid='agusza'
In your query below you doesnt state the database table where you should get that data using FROM
$query = "select * where userid = 'agusza' "; // instead of this
$query = "select * FROM declaredtable where userid = 'agusza' "; used this
I have been searching around for about 30 minutes, and I just can't find the issue here...
if($middle == 'garage'){
$result = mysql_query("SELECT * FROM character WHERE username = '$username'") or die(mysql_error());
$row = mysql_fetch_array($result);
$return['middle'] = $row['username'];
}
For some reason this is returning
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 'character WHERE username = 'Alcapwn'' at line 1
I have tried so many things, other SELECT * FROM query's are working fine, it's just this one. The table and row containing it is there. If I switch
$result = mysql_query("SELECT * FROM character WHERE username = '$username'") or die(mysql_error());
to
$result = mysql_query("SELECT * FROM users WHERE username = '$username'") or die(mysql_error());
the second one will work, but not the first.
character is a reserved mysql word, so you have to enclose it in backticks
FROM `character` ...
It seems that character is a MySQL Reserved Keyword, which means you cannot use it as table name.
"Character" is a reserved word in MySQL.
http://dev.mysql.com/doc/refman/5.0/en/reserved-words.html
Encountered with the following warning when trying to access the page:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/designand/www/www/random.php on line 13
Everything was working fine when I was testing it on XAMPP.
<?php
$db_hostname = "localhost";
$db_username = "root";
$db_name = "links";
$db_pass = "xxx";
$dbh = mysql_connect ($db_hostname, $db_username, $db_pass) or die ('I cannot connect to the database because: ' . mysql_error());
mysql_select_db($db_name) or die(mysql_error());
$num_displayed = 1 ;
$result = mysql_query ("SELECT * FROM 'links' ORDER BY RAND() LIMIT $num_displayed");
while($row = mysql_fetch_array( $result ))
{
echo "<img src=\"" . $row["image"] . "\" border=0 alt=\"\">" ;
}
mysql_close($dbh);
?>
An error like that almost always means there's a problem with your query.
To find out what error message MySQL returns, you can put or die(mysql_error()) just before the semicolon after your mysql_query call. You may also want to print the exact query text you send to MySQL, as this may help you to see the actual problem.
Given that I don't know how much you may have anonymized this example, I can't be sure if this is the actual error, but it looks like you've surrounded your table name with apostrophes ('). This is incorrect; the correct character for escaping table and column names is `.
"supplied argument is not a valid MySQL result resource" means that the query didn't return a valid resource. It failed. To view the error just print out mysql_error() after mysql_query().
Could be that the table doesn't exist. Did you check it?
The second thinkg - ORDER BY RAND() is bad! You should think of different ways on how to shuffle the result. Just google it, there are a lot of other ways to do it - ORDER BY RAND() # Google
Try changing this line:
$result = mysql_query ("SELECT * FROM 'links' ORDER BY RAND() LIMIT $num_displayed");
To:
$result = mysql_query ("SELECT * FROM 'links' ORDER BY RAND() LIMIT $num_displayed") or die (echo mysql_error());
It seems like the SQL is failing to return a valid response. Adding the above should show you the last MySQL error and help point you in the correct direction.
Please add this code to get if the mysql is throwing any errors:
$result = mysql_query('SELECT * WHERE 1=1');
if (!$result) {
die('Invalid query: ' . mysql_error());
}
while(....