Im trying to run a MySQL query, but for some reason its not working.
This is the query im running:
SELECT * FROM applications WHERE status = 0
And my database looks like this:
http://gyazo.com/cc0bfa109e73a771d99a22b5051ee2de
However, num_rows() returns 0 rows...
No errors are displayed...
Try below code hope it will help you because i have tested in localhost.
$sql = "SELECT * FROM applications WHERE status = 0";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result))
{
echo $row['mcUser'];
}
IF its not work's check your table name again.
replace the query with single quotes between 0
$sql = "SELECT * FROM applications WHERE status = '0'";
Related
ok, so I have a search script with is sort of working but I dont get the results I want.
when i run the query in php it doesnt result anything.
$searchStr = htmlspecialchars($searchStr);
$sql = "SELECT * FROM steamitems WHERE steamid='" . $id . "' AND name LIKE '%".$searchStr."%'";
$r_query = mysqli_query($link, $sql);
but if I run the exact same output as $sql (SELECT * FROM steamitems WHERE steamid='76561198196240283' AND name LIKE '%sawed%') in phpmyadmin it returns the correct result..
EDIT: I forgot to mention that I obviously print the results here
while ($row = mysqli_fetch_array($r_query)){
echo $row["assetid"];
echo $row["name];
}
I have this mysql query in my mysql include file sql.php :
$resultid = array();
$sqlstring9 = "SELECT * FROM auswahl_time WHERE IsActive = 1 ORDER BY Time_PKey ASC";
$resultid[9] = mysql_query($sqlstring9);
$resultid10 = mysql_query($sqlstring9);
On my index page :
echo mysql_num_rows($resultid[9]);
echo mysql_num_rows($resultid10);
returns only the result for $resultid10 not $resultid[9]. I got this to work before and im not sure what changed since then. What are the requirements for this to work? Do i need to initialize the array for resultid in every script?
I found the problem, turns out i had another unrelated query called $resultid. After renaming it it worked.
Try this
$sqlstring9 = "SELECT * FROM auswahl_time WHERE IsActive = 1 ORDER BY Time_PKey ASC";
$resultid = mysql_fetch_assoc(mysql_query($sqlstring9));
echo $resultid[9];
echo $resultid[10];
I am making a page that queries a table for all columns of all results ordered by entry time in descending order with a limit. When I query for a count of the rows, the query works just fine, but when I try to query the table again for data, I don't get anything. I decided to try cutting the query down to "SELECT * FROM comments" but I still got no results when "SELECT COUNT(*) AS count FROM comments" just beforehand worked. I've tried using mysqli_error(), but that didn't give me any information.
The query doesn't seem to be failing as the result from mysqli_query() isn't false and when I query in phpMyAdmin, the queries work. A little piece of my code below
//open databases
require_once($root . "databases/data.php");
//get number of suggestions in comments table
$cquery = mysqli_query($cbase, "SELECT COUNT(*) AS count FROM comments"); //this works
$c = mysqli_fetch_array($cquery);
$count = $c["count"];
//get all suggestions
//this query fails
$queryText = "SELECT * FROM comments ORDER BY time DESC LIMIT " . (($page - 1) * $pageLimit) . ", " . $pageLimit;
$query = mysqli_query($cbase, $queryText);
//validate query
if($query === false)
{
$failed = true;
}
//get all comments from query
while(!$failed && $array = mysqli_fetch_array($result))
Please try this on line 3
$c = mysqli_fetch_assoc($cquery);
You can also try like this also,
$c = mysqli_fetch_array($cquery, MYSQLI_ASSOC);
You are just using the wrong variable when reading out your query results in your while-loop. mysqli_fetch_array($result) while you saved the query-result in $query so it should be mysqli_fetch_array($query)
Was hoping someone could give me some help. I have been trying to learn pagination but have run into an issue. when I run this to get my total rows count:
$sql = "SELECT COUNT (*) FROM item WHERE fid='17'";
$query = mysqli_query($con, $sql);
$row = mysqli_fetch_row($query);
$rows = $row[0];
$rows comes back with no value, I am under the impression that $rows should contain the total number of records from item with the fid of 17, when I run SELECT COUNT (*) FROM item WHERE fid='17' in phpmyadmin it returns 98 which is the correct count. Directly before the above code I use this code to connect to the db, which I use again later to display the records which works fine.
$con=mysqli_connect("$host","$username","$password","$dbname");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
This statement displays records later in the script
$sql = mysqli_query($con,"SELECT * FROM item WHERE fid='17' ORDER BY id DESC $limit ");
So there is data and the info is correct.
I have been following this tutorial on the subject http://www.developphp.com/view.php?tid=1349 and it works like a charm on his example in the video, all I have changed is the database particulars to mine. Can't figure this one out, been stuck for days so now I am bothering you fine folks.
Update: Changed
$query = mysqli_query($con, $sql);
$row = mysqli_fetch_row($query);
$rows = $row[0];
to
if ($result=mysqli_query($con,$sql))
{
// Return the number of rows in result set
$rows=mysqli_num_rows($result);
// Free result set
mysqli_free_result($result);
}
And everything is working now. Still wish I knew what was wrong with my original code but I will be moving on. Thanks Phil Perry for all your help this morning!
Well, you could SELECT * FROM... and then call mysqli_num_rows($query). Probably slower than SELECT count(*). BTW, I presume that item is not a reserved word in SQL. You can always wrap table and field names in backticks (not quotes ' or "). You could also try SELECT count(*) AS myCount....
try this,
$sql = "SELECT COUNT(*) FROM item WHERE fid='17'";
$query = mysqli_query($sql);
$row = mysqli_fetch_array($query);
$rows = $row[0];
I'm trying to display a field from my MySQL database. It's in the table tblproducts in the row with the id is set to 1. And under the column qty.
This is the code I'm using:
<?php
mysql_connect("localhost","username","password");
mysql_select_db("database_name");
$available = "SELECT qty FROM tblproducts WHERE id = 1";
$result = mysql_query($available);
echo $result;
?>
However, I keep getting this message: Resource id #2
I've done a bit of research and seen where other people are having similar problems but most of them are trying to display their data in an HTML table whereas I just need the data from 'qty' to display. And of course I'm definitely not a MySQL guru.
Can anyone help me out with this please?
Try changing this:
$result = mysql_query($available);
To this:
$result = mysql_result(mysql_query($available), 0);
Let's start from the start. (I'll assume you have the connection set)
Form the query
$query = "SELECT `qty`
FROM `tblproducts`
WHERE `id` = 1";
Execute the query
$run = mysql_query($query);
Now, put the result in an assoc array
$r = mysql_fetch_array($run);
See the contents of the array
echo $r['qty'];
It's also advised that you move up from mysql to either mysqli, or PDO. PDO is preferred as you're not bound to the MySQL database model.
Try this:
Here you need to generate associative array and then get the resulting row.
$query = "SELECT `qty` FROM `tblproducts` WHERE `id` = 1";
$run = mysql_query($query);
$r = mysql_fetch_array($run);
echo $r['qty'];
-
Thanks