I am getting this error from the following code:
function updateRank($master_list, $event_id)
{
$cnt_ml = count($master_list);
echo "count master list = $cnt_ml<br>";
$b=1;
for ($k=0; $k<$cnt_ml; $k++)
{
echo "master list element 1 - ".$master_list[1]."<br>";
$foo = $master_list[$k];
// Update each team in event_team table
$update = "UPDATE event_team
SET pool_rank = $b
WHERE event_id = $event_id
AND team_id = $foo";
mysqli_query($conn, $update);// or die ('Could not run insert in event_team table');
echo "|".$update."|<br>"; // Leave this line for debugging the sql query.
$b++;
}
}
Specifically "$foo" is throwing the error. The "master_list" array is being passed correctly. When running the code, $cnt_ml returns "5" (which is absolutely correct). The line with:
echo "master list element 1 - ".$master_list[1]."<br>";
Returns "54" (which is correct).
Since the script is successfully able to read the array elements and post them, why is the script throwing the error when I include it in the UPDATE? It is able to see the array data but choking when trying to use the data in the UPDATE.
Thanks in advance!
Variable $conn is not declared. Or it's declared globally and you forgot to add
global $conn;
in updateRank function
Related
I have a PHP class where I am initializing an array retrieved from database in the constructor:
class TableClass{
public function __construct($con, $id){
$this->con = $con;
$query = mysqli_query($this->con, "SELECT * FROM table1 WHERE id='$id'");
$this->table_result= mysqli_fetch_array($query);
}
}
I have different functions that I want to use the array without having to fetch again. I just want to loop through the results and make some calculations.
I tried the following syntax:
public function getNumberOfComments(){
for ($i = 0; $i < count($this->table_result); $i++) {
$comment= $this->table_result[$i]['comment'];
}
}
I am getting an error:
Illegal string offset "comment"
What is the correct syntax?
You are fetching one row from the result set:
$this->table_result= mysqli_fetch_array($query);
Assuming that the id is the primary key, you will have 0 or 1 rows in your result set, so if a row is found, you can access the fields directly without using a loop:
$comment= $this->table_result['comment'];
Illegal string offset "comment" is telling you that the value of $this->table_result[$i] is not an array, and therefore does not contain an element named comment.
The problem is possibly being caused by the code in the constructor failing to work properly; you don't have any error checking in there, so if it fails, it won't tell you about it.
Add some error checking in the constructor, find out why it's failing, and fix it.
When I run this:
echo $user_id; // prints 3
$tags_ids_results = $mysqli->query("SELECT TagID FROM UserTagSubscriptions WHERE UserID = $user_id");
while($tag_id_row = $tags_ids_results->fetch_object()) {
$tag_id = $tag_id_row->TagID;
echo "In While Loop";
}
This only prints "In While Loop" once. However, when I run the query against the database directly (with the same value of $user_id), three rows are returned.
UPDATE
When I call fetch_all() on $tags_ids_results, I get a call to undefined method error (I'm on PHP 5.4). I also get NULL when I print out num_row.
Turns out I was redeclaring $tags_ids_results in the rest of my loop.
I'm having trouble getting info from my MySQL database.
Here is my code :
/********************
* Database Info
********************/
$host = "localhost";
$user = "admin";
$pass = "admin#";
$database = "db_admin";
/********************
* Database connection
********************/
$con = mysqli_connect( $host, $user, $pass, $database );
if (mysqli_connect_errno ()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error ();
}
$result = array();
if (isset($_POST['ID'])) {
$id = $_POST['ID'];
$query = "SELECT * FROM Servers WHERE PID='" .$id. "'";
$result = mysqli_query($con, $query);
}
print("<pre>".print_r($result,true)."</pre>");
My first question, Did I use "isset" function currectly?
Because it doesnt seem like it is actually going though the if statement.
The Url I am using is : #..com/view.php?ID=1
My second question, Did I use the $query correctly?
Because I echo $id and that echoed out a "MySQL Object()"
Finally, the print printed out "Array()"
I'm just starting on PHP, Thanks for the help :)
A few things:
If you're passing the variable in the query string, use $_GET instead of $_POST to retrieve the values.
$result will return an pointer to the recordset, not the rows themselves. You will have to use mysqli_fetch_array() to fetch the rows.
ADD:
If you are sure that you will only have 1 record returning, you can use:
$row = mysqli_fetch_assoc($query);
echo $row['field_name'];
More then 1 record?
while($row = mysqli_fetch_assoc($query)){
echo $row['field_name'];
}
# your first question: if you have a input field with the name="ID", then its good.
Please also post your HTML :)
$var = 'Hello world';
if(isset($var)){ //If the var $var has been set (in this case it is)
echo $var;
} else {
//If $var is not set, then we get in the else
echo 'The var $var is not set';
}
The best thing is debugging the code with a debugger, you may use XDebug, or at least use var_dump(); to see what happens
var_dump($_REQUEST, $result);
Answer to your first question: It's hard to say if you've used it correctly when you haven't said what you're trying to do. I'm presuming that you only want run the code enclosed in the if-statement if the POST variable 'ID' has been received. If so, yes you've done it correctly.
Answer to your second question: I'm presuming on this line you're trying to build a string with a valid MySQL query. You've done that correctly, assuming $_POST['ID'] is a string (or can be converted to a string, see http://www.php.net/manual/en/language.types.string.php#language.types.string.casting).
If you're echoing $id and it's returning an object, however, you'll have a problem. You can't combine a string and an object like that. You'd need to iterate the object with a foreach, for example, and extract the id from that. The rest of the code won't work until that part is resolved.
The thing to investigate now is why $_POST['ID'] is returning an object. You'll need to provide the form code at the very least.
I have a select statement where I want to get all rows from a table but seem to be having a mental blockage - this should be elementary stuff but can't seem to get it working.
There are only two rows in the table 'postage_price' - and two columns : price | ref
Select statement is as follows:
$get_postage="SELECT price FROM postage_price ORDER BY ref DESC";
$get_postage_result=mysqli_query($dbc, $get_postage) or die("Could not get postage");
while($post_row=mysqli_fetch_array($dbc, $get_postage_result))
{
$post1[]=$post_row;
}
I am then trying to echo the results out:
echo $post1['0'];
echo $post1['1'];
this is not showing anything. My headache doesn't help either.
while($post_row = mysqli_fetch_array($dbc, $get_postage_result))
{
$post1[] = $post_row['price'];
}
As you see: $post_row in this line: = mysqli_fetch_array($dbc, $get_postage_result) is an array. You are trying to save the whole array value to another array in a block. :)
EDIT
while($post_row = mysqli_fetch_array($get_postage_result))
...
You have $post1[]=$post_row; and $post_row is itself an array. So you can access post data with following: $post1[NUMBER][0] where NUMBER is a $post1 array index and [0] is 0-index of $post_row returned by mysqli_fetch_array.
Probably you wanted to use $post1[]=$post_row[0]; in your code to avoid having array of arrays.
You are passing 1 and 0 as string indexes, this would only work if you had a column called 0 or 1 in you database. You need to pass them as numeric indexes.
Try:
print_r($post1[0]);
print_r($post1[1]);
or
print_r($post['price']);
print_r($post['ref']);
with all your help I have found the error - it is in the mysqli_fetch_array where I had the $dbc which is not required.
$get_postage="SELECT price FROM postage_price ORDER BY ref DESC";
$get_postage_result=mysqli_query($dbc, $get_postage) or die("Could not get postage");
while($post_row=mysqli_fetch_array($get_postage_result))
{
$post1[]=$post_row['price'];
}
instead of:
$get_postage="SELECT price FROM postage_price ORDER BY ref DESC";
$get_postage_result=mysqli_query($dbc, $get_postage) or die("Could not get postage");
while($post_row=mysqli_fetch_array($dbc, $get_postage_result))
{
$post1[]=$post_row['price'];
}
Bad day for me :(
Thanks all
If something does not work in a PHP script, first thing you can do is to gain more knowledge. You have written that
echo $post1['0'];
echo $post1['1'];
Is showing nothing. That could only be the case if those values are NULL, FALSE or an empty string.
So next step would be to either look into $post1 first
var_dump($post1);
by dumping the variable.
The other step is that you enable error display and reporting to the highest level on top of your script so you get into the know where potential issues are:
ini_set('display_errors', 1); error_reporting(~0);
Also you could use PHP 5.4 (the first part works with the old current PHP 5.3 as well, the foreach does not but you could make query() return something that does) and simplify your script a little, like so:
class MyDB extends mysqli
{
private $throwOnError = true; # That is the die() style you do.
public function query($query, $resultmode = MYSQLI_STORE_RESULT) {
$result = parent::query($query, $resultmode);
if (!$result && $this->throwOnError) {
throw new RuntimeException(sprintf('Query "%s" failed: (#%d) %s', $query, $this->errno, $this->error));
}
return $result;
}
}
$connection = new MyDB('localhost', 'testuser', 'test', 'test');
$query = 'SELECT `option` FROM config';
$result = $connection->query($query);
foreach ($result as $row) {
var_dump($row);
}
I have a php variable: $foo
My MySQL table called data has the following structure:
id var header
1 zj3 http://google.com
I would like to check if $foo is all ready in var row.
If it is I would like to echo header ("http://google.com")
How would you approach this?
Thanks in advance, please ask if any clarification is needed!
Your query should be:
SELECT `header` FROM `data` WHERE `var` = '$foo'
This will return all the headers with a var value of $foo.
$db = mysqli_connect('localhost', 'username', 'password', 'database');
if($query = mysqli_query($db, "SELECT `header` FROM `data` WHERE `var` = '$foo'")){
while($row = mysqli_fetch_assoc($query)){
echo $row['header'];
}
mysqli_free_result($query);
}
first connect to the db
$query = mysql_query("SELECT var, header FROM data WHERE id='1'") or die(mysql_error());
while($row = mysql_fetch_assoc($query)){
if($foo == $row['var']){
echo $row['header'];
}
}
EDIT: changed equality statement based on your edit
It's not difficult at all, If I understand correctly then this should help you.
// Query Variable / Contains you database query information
$results = $query;
// Loop through like so if the results are returned as an array
foreach($results as $result)
{
if(!$result['var'])
echo $result['header'];
}
// Loop through like so if the results are returned as an object
foreach($results as $result)
{
if(!$result->var)
echo $result->header;
}
are you asking if $foo matches any of the fields in data, or if $foo=some_field? Here for if you want $foo==var.
$foo='somevalue';
$query="SELECT id, var, header FROM `data` WHERE var='$foo'";
$result=mysqli_query($query);
if($result->num_rows==0)
$loc= 'http://google.com';//default value for when there is no row that matches $foo
}else{
$row=$result->fetch_assoc(); //more than one row is useless since the first header('Location: x') command sends the browser to a new page and away from your script.
$loc=$row['header'];
}
header ("Location: $loc);
exit;
ETA: since you've edited your question, it appears that you want to echo the header column if your search value matches your var column. The above won't work for that.
You just want to know if $var's value is anywhere in that column (any row(s))?
SELECT COUNT(id) FROM data WHERE var = ?;
The result will be the number of rows for which the field var contains the value of $var.
Here's a template for all the "does it exist" questions.
This is the only thing that actually worked for me so far and is not deprecated.
if ($query = mysqli_query($link, "SELECT header FROM data WHERE var = '$foo'")) {
$header = mysqli_fetch_assoc($query);
if ($header) {
// The variable with value $foo exists.
}
else {
// The variable with value $foo doesn't exist.
}
}
else {
// The query didn't execute for some reason. (Dammit Obama!)
}
WARNING!
Even if the variable DOES NOT EXIST the comparison between $query and mysqli_query() will always return TRUE.
The only way --which happened to me-- for the comparison to return FALSE is because of a syntax error in your query.
I don't know why it worked for the guy who wrote the accepted answer, maybe it's an update or maybe he had a syntax error and was so confident that he didn't check if it could ever be TRUE.
Here's the comment someone made for correcting his syntax:
"Add another ) before the { in the first line"
So, the accepted answer is WRONG!