I have never used JOINs or have worked with multiple table before.
This error is popping up.
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in [pathway to this mysql_query line].
What am I doing wrong here?
Thank you.
$group_id= 4;
$result = mysql_query("SELECT table1.user_facebook_id AS user_facebook_id
FROM table1 JOIN table2
ON table1.user_id = table2.user_id
WHERE table2.group_id = $group_id");
$row = mysql_fetch_assoc($result);
You should test the value of $result to see if the query failed. If so, print the error for debugging purposes:
$sql = "SELECT ...";
$result = mysql_query($sql);
if (!$result) {
trigger_error(mysql_error());
}
You may also want to try running the SQL query in the MySQL workbench to see if it works there.
Related
I'm querying a server and iterating through multiple databases using PHP, but for some reason this $sql2 query (which I have read works in countless threads) is returning a syntax error:
$res = mysqli_query($conn,"SHOW DATABASES");
if (!$res){
// Deal with error
}
while ($d = mysqli_fetch_array($res)){
$db = $d['Database'];
$sql1 = "USE $db";
$query1 = mysqli_query($conn, $sql1);
if (!$query1){
// Deal with error
}
$sql2 = "IF (EXISTS (SELECT *
FROM INFORMATION_SCHEMA.TABLE
WHERE TABLE_SCHEMA = '$db'
AND TABLE_NAME = 'appusers'))
BEGIN
SELECT * FROM `appusers`
END";
$query2 = mysqli_query($conn, $sql2);
if (!$query2){
// Deal with error
}
}
This is the error I receive:
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 'IF (EXISTS (SELECT * FROM INFORMATION_SCHEMA.TABLE WHERE TABLE_S' at line 1
My MySQL Server version is 5.6.27 and my PHP interpreter is 5.6
You can't use an IF statement as a query, only in a stored procedure. You'll need to perform two separate queries.
$sql = "SELECT COUNT(*) AS count
FROM INFORMATION_SCHEMA.TABLES
WHERE TABLE_SCHEMA = '$db'
AND TABLE_NAME = 'appusers'";
$result = mysqli_query($conn, $sql) or die(mysqli_error($conn);
$row = mysqli_fetch_assoc($result);
if ($row['count'] != 0) {
$sql2 = "SELECT * FROM appusers";
$query2 = mysqli_query($conn, $sql2);
...
} else {
// deal with error
}
There is no if in SQL (although mysql has a function called if)
Apparently, you want to run a query if the table exists. The more common way would be to try running the query and check whether the error you get says that the table doesn't exist.
E.g. if you run the query in the mysql command line application, you might get this:
mysql> SELECT * FROM appusers;
ERROR 1146 (42S02): Table 'test.appusers' doesn't exist
You see two codes:
1146: internal mysql code (obtain this via mysqli_errno)
42S02: the SQL standard error state (obtain this via mysqli_sqlstate)
In your case, checking the SQL state might be better because it covers more cases. E.g., all of these mysql error codes map to SQL state 42S02:
1051 - Unknown table '%s'
1109 - Unknown table '%s' in %s
1146 - Table '%s.%s' doesn't exist
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'm trying to pull some information from a database, and the connection is working, but for some reason it isn't recognizing my query, even though I confirmed the query in the database with SQL and had it "generate PHP code". The echo statement is coming up blank. It's a mySQL database. Thanks for your help.
$query = "SELECT `contact` FROM `contactinfo` WHERE member=\'Henry\'";
$contact = mysqli_query($db,$query);
echo $contact;
$contact contains MySQL result object you need to fetch data from this to use this in your application.
$query = "SELECT `contact` FROM `contactinfo` WHERE member = 'Henry'";
$contact = mysqli_query($db, $query);
while ($row = mysqli_fetch_row($contact)) {
echo $row[0]; // 0 to n indicates the Column(s) Selected in SELECT Query
}
All,
If I run a query like the following:
$qry = "Select wrong_column from table_name";
$result = mysql_query($qry);
If wrong_column doesn't exist then I'll get a mySQL error. In PHP, how can I determine if there was an error from mySQL? If there was an error I'd like it to stop further processing but if there wasn't an error I'd like to get the mySQL results like this:
$resultset = mysql_fetch_array($result);
Would doing something like this work?
if(!mysql_error()){
$resultset = mysql_fetch_array($result);
}
Any advice on how to do this would be appreciated. Thanks in advance!
how can I determine if there was an error from mySQL?
To see if an error occured you should test the result of mysql_query:
$result = mysql_query($qry);
if (!$result) {
$error = mysql_error();
// Handle the error.
}
From the documentation:
For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.
Emphasis mine.
mysql_query("SELECT foo FROM bar") or exit(mysql_error()); is your best friend =)
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
I' really stuck on this , I'm gettiing this error:
mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in "filename"
Here is the code:
$sql = "SELECT * FROM $tbl_name WHERE....
$result=mysql_query($sql);
$row = mysql_fetch_assoc($result);
The wierd thing is that I've used the exact same code before and it worked fine
Any ideas??
That means the query failed. Usually it's a SQL syntax error. To find out, just insert this right before the _fetch_assoc line:
print mysql_error();
To prevent the error message, structure your code like this to check the $result beforehand:
$sql = "SELECT * FROM $tbl_name WHERE....";
if ($result = mysql_query($sql)) {
$row = mysql_fetch_assoc($result);
}
else print mysql_error();
Always run all your queries this way
$sql = "SELECT * FROM $tbl_name WHERE....";
$res = mysql_query($sql) or trigger_error(mysql_error()." in ".$sql);
$row = mysql_fetch_assoc($result);
And you will be notified of the cause of error.
But never print or let die() output any errors, as it's security flaw.
This error usually occurs because no data is returned from the query. Make sure the data is being returned by going into something like PHPMyAdmin and making sure the query returns some rows. You should also add the
or die(mysql_error());
At the end of your query.