Trying to display a message if no results found in the search query.
Here is a brief example of my code.
$sql = "SELECT * FROM details WHERE ID =1"
$res =& $db->query($sql);
if (PEAR::isError($res)) {
die($res->getMessage());
}
while($row = $res->fetchRow())
{
echo 'results'
{
Any help would be greatly appreciated! Thanks
DB_result has a method called numRows(), so you could check
if($res->numRows() == 0)
http://pear.php.net/package/DB/docs/latest/DB/DB_result.html#methodnumRows
Get the number of rows in a result set
Return: the number of rows. A DB_Error object on failure.
And you don't need to worry about DB_Error since it was already checked at
PEAR::isError($res)
Related
I'm sorry to have to post this, I know it seems similar to other posts, so I ask you to please bear with me...
What I am trying to do: I have a sql query that is in a function, the point of the function is to find if there are TinyInts of a particular value (1 or 0), I am doing a count, and need to just return effectively if it found the value in the specified in the query. What should be coming back is 1 but I only get 0's back...
Here is the whole function:
//return count of VALUES
function getcountsroles($searchParams, $searchCrit){
/// does query, gets results, counts the number and returns that true or false
///this willbe used to show the dashboard user control more efficently.
// $searchParams == column
// $searchCrit == int 1 or 0
//should fetch back the count back from the query
global $db, $dashboard_message_users;
$count_Data=array();
$query = "SELECT COUNT(userID) AS countVal FROM roles WHERE ? = ?";
$stmnt = $db -> prepare($query);
if(!$stmnt){
echo var_dump($stmnt);
}
$stmnt -> bind_param("si", $searchParams, $searchCrit);
if(!$stmnt -> execute()){
$dashboard_message_users = "<p class='alert alert-danger'>The db query faulted.</p>";
}
$result = $stmnt -> get_result();
while ($data = $result->fetch_assoc()){
$count_Data[] = $data;
}
$test = 'userID';
$value = (int)($data['countVal']);
echo var_dump($test);
echo var_dump($searchCrit);
echo var_dump($searchParams);
echo var_dump($value);
echo var_dump($count_Data);
if ($value > 0){
return true;
}else{
return false;
}
}
The calling function sends the column to count the values of, and the the value that we are looking for. when I run the query against the database I get the correct values, I have tested with different columns and values (again 1 or 0)
SELECT COUNT(*) as 'countVal' FROM roles WHERE isUser = 1;
SELECT * FROM roles WHERE isUser = 1;
SELECT * FROM roles;
SELECT COALESCE(COUNT(isUser),0) AS totid FROM roles;
the first one returns 2, the second is all the rows where isUser is value of 1,
third grabs all rows... and the last one was a attempt to see if it was going work... but it still only returns 0 to the result... I have also tried to get the count of rows should be 1, either if true or false... This i getting annoying, Unfortunately I happen to be using mysqli lib and most exampled are for POD or old mysql_PHP, and in OO it seems to not like the prepared statements.
I am positive its something I am doing, I have looked over and over my code and I cant see what the issue is... Maybe if a fresh pair of eyes.
Any time or replies that help me and others out would be greatly appreciated.
Jesse Fender
So I am using this tutorial: https://www.simplifiedcoding.net/android-mysql-tutorial-to-perform-basic-crud-operation/ to try and get data from my local MYSQL server (using Wamp64). I had the undefined index error at first, which I fixed using the isset() statement.
But now it just returns:
{"result":[]}
I have, however, a lot of data in the set column of that database.
Here is the code:
<?php
//Getting the requested klas
$klas = isset($_GET['klas']) ? $_GET['klas'] : '';
//Importing database
require_once('dbConnect.php');
//Creating SQL query with where clause to get a specific klas
$sql = "SELECT * FROM lessen WHERE klas='$klas'";
//Getting result
$r = mysqli_query($con,$sql);
//Pushing result to an array
$result = array();
while ($row = mysqli_fetch_array($r)) {
array_push($result,array(
"id"=>$row['id'],
"klas"=>$row['klas'],
"dag"=>$row['dag'],
"lesuur"=>$row['lesuur'],
"les"=>$row['les'],
"lokaal"=>$row['lokaal']
));
}
//Displaying the array in JSON format
echo json_encode(array('result'=>$result));
mysqli_close($con);
?>
I tried out the
SELECT * FROM lessen WHERE klas='$klas'
statement in my database and it seems to return the correct data.
Any idea what is causing this?
Thanks in advance!
Point 1 is:
isset function only checks if klas is set in the $_GET global array. So if somehow $klas is blank - your query will return empty (without giving error).
So please check values in the $_GET and possibly from where it is accessed. Or you can add condition to avoid empty query like --
if (!empty($_GET['klas'])) {
// rest of the code block upto return
Point 2 is:
You have mentioned if you echo the sql it returns
SELECT * FROM lessen WHERE klas=''{"result":[]}
Here the second part (the JSON) is from echoing the result at the end of your code. So for the first part (i.e. echoing $sql) we see that klas=''. That actually goes to the Point 1 as mentioned above.
So finally you have to check why the value at $_GET is showing blank. That will solve your problem.
UPDATE:
From #GeeSplit's comment For the request
"GET /JSON-parsing/getKlas.php?=3ECA"
There will be nothing in $_GET['klas'] cause the querystring in the url doesn't contain any key.
So either you have to change the source from where the file is called. Or you can change how you are getting the value of klas.
Example:
$tmpKlas = $_SERVER['QUERY_STRING'];
$klas = ltrim($tmpKlas, '=');
Rest of your code will work.
Use this code
<?php
$klas ='';
if(isset($_GET['klas']) && !empty($_GET['klas']))
{
$klas = $_GET['klas'];
require_once('dbConnect.php');
$sql = 'SELECT * FROM lessen WHERE klas="'.$klas.'"';
$r = mysqli_query($con,$sql);
$result = array();
while ($row = mysqli_fetch_array($r)) {
array_push($result,array(
"id"=>$row['id'],
"klas"=>$row['klas'],
"dag"=>$row['dag'],
"lesuur"=>$row['lesuur'],
"les"=>$row['les'],
"lokaal"=>$row['lokaal']
));
}
echo json_encode(array('result'=>$result));
mysqli_close($con);
}
?>
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!
I tried what seemed like the most intuitive approach
$query = "SELECT * FROM members
WHERE username = '$_CLEAN[username]'
AND password = '$_CLEAN[password]'";
$result = mysql_query($query);
if ($result)
{ ...
but that didn't work because mysql_query returns a true value even if 0 rows are returned.
I basically want to perform the logic in that condition only if a row is returned.
Use mysql_num_rows:
if (mysql_num_rows($result)) {
//do stuff
}
If you're checking for exactly one row:
if ($Row = mysql_fetch_object($result)) {
// do stuff
}
You can use mysql_fetch_array() instead, or whatever, but the principle is the same. If you're doing expecting 1 or more rows:
while ($Row = mysql_fetch_object($result)) {
// do stuff
}
This will loop until it runs out of rows, at which point it'll continue on.
mysql_num_rows
Retrieves the number of rows from a result set. This command is only valid for statements like SELECT or SHOW that return an actual result set.
If none match, then zero will be the return value and effectively FALSE.
$result = mysql_query($query);
if(mysql_num_rows($result))
{ //-- non-empty rows found fitting your SQL query
while($row = mysql_fetch_array($result))
{//-- loop through the rows,
//-- each time resetting an array, $row, with the values
}
}
Which is all good and fine if you only pull out of the database. If you change or delete rows from the database and want to know how many were affected by it...
To retrieve the number of rows affected by a INSERT, UPDATE, REPLACE or DELETE query, use mysql_affected_rows().
$result = mysql_query($query);
if(mysql_affected_rows())
{ //-- database has been changed
}
//-- if you want to know how many rows were affected:
echo 'Rows affected by last SQL query: ' .mysql_affected_rows();
mysql_query() will only return FALSE if the query failed. It will return TRUE even if you have no rows, but successfully queried the database.
$sql = "SELECT columns FROM table";
$results = mysql_query($sql, $conn);
$nResults = mysql_num_rows($results);
if ($nResults > 0) {
//Hurray
} else {
//Nah
}
This should work.
I used the following:
if ($result != 0 && mysql_num_rows($result)) {
If a query returns nothing it will be a boolean result and it's value will be 0.
So you check if it's a zero or not, and if not, we know there's something in there..
HOWEVER, sometimes it'll return a 1, even when there is nothing in there, so you THEN check if there are any rows and if there is a full row in there, you know for sure that a result has been returned.
What about this way:
$query = "SELECT * FROM members WHERE username = '$_CLEAN[username]'
AND password = '$_CLEAN[password]'";
$result = mysql_query($query);
$result = mysql_fetch_array($result);
//you could then define your variables like:
$username = $result['username'];
$password = $result['password'];
if ($result)
{ ...
I like it because I get to be very specific with the results returned from the mysql_query.
-Ivan Novak
well...
by definiton mysql_query:
mysql_query() returns a resource on
success, or FALSE on error.
but what you need to understand is if this function returns a value different than FALSE the query has been ran without problems (correct syntax, connect still alive,etc.) but this doesnt mean you query is returning some row.
for example
<?php
$result = mysql_query("SELECT * FROM a WHERE 1 = 0");
print_r($result); // => true
?>
so if you get FALSE you can use
mysql_errorno() and mysql_error() to know what happened..
following with this:
you can use mysql_fetch_array() to get row by row from a query
$result = mysql_query(...);
if(false !== $result)
{
//...
}