Im starting to use sqlite3 but I'am unable to get some results.
My database is called database.sqlite3
Inside SQLite command prompt and doing 'SELECT * FROM table'; . works fine. But im having problems with php.
$db = new SQLite3('db/database.sqlite3');
$result = $db->query('SELECT * FROM table');
var_dump( $result );
The result of this code is : object(SQLite3Result)#3 (0) { }
What is worng here?
Thank you.
PD: $result->fetchArray(SQLITE3_ASSOC) works but only gives me one record... why???
from the php doc, the return value of query:
Return Values
Returns an SQLite3Result object, or FALSE on failure.
Have a look at the fetcharray method of the Result object to see how to get (fetch) the the row data.
So just in case someone is in the same trouble. SQLITE does not work like that. You must walk through a loop to get every row. Then you can create your data array.
Something like this.
$db = new SQLite3('db/database.sqlite3');
$result = $db->query('SELECT * FROM table');
while( $row = $result->fetchArray() ){
echo $row['name'];
}
Related
I'm updating some old code that has deprecated MySQL functions. But for some reasons I cannot get all the results from the column. The strange part is that if I run the query directly on the server I get all results fine. So this is an issue with PHP getting the results, not the MySQL server or my query.
Here is the new and old code:
My current updated code:
$sql = "SELECT user, monitor FROM users WHERE `status` = 'y'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
// This works. It shows all results
echo $row["user"];
// This does not work! Only shows one result:
$account= $row["user"];
}
else {
echo 'No results';
}
When I use that query directly on DB server, I get all results. So the SQL query is correct. I actually also get all results as well in PHP if I echo the row directly like:
echo $row["user"];
But for some reason when I try to use it with a PHP with variable it only lists one user result.
In the past I used this but the mysql_fetch_array function is now deprecated
while ($row = mysql_fetch_array($result)) {
array_push($data, $row["user"]);
}
foreach($data as $value) {
$account = $value
}
I cannot use my previous code anymore as those MySQL functions are obsolete today. I need to write the results into a file and my old method worked fine. The new one using mysqli does not.
Any suggestions?
You just need to add one of these [, and one of these ].
$account[] = $row["user"];
// ^^ right here.
$account= $row["user"]; means you're storing the value of $row["user"] in $account each time the loop executes. $account is a string, and it gets a new value each time.
$account[] = $row["user"]; means you're appending each value of $row["user"] to an array instead.
You should not use array_push for this. It's overkill for appending a single value to an array. And if the array isn't defined beforehand, it won't work at all.
I am terrible at PHP and I need to retrieve data from a database and give it to an index.php view. The view is pre-made and has this code:
//This is simplified - it has error handling that is not shown
$results = getAll($tableName);
//This is the line where it is failing
//Undefined Offset
$columns = empty($results) ? array() : array_keys($results[0]);
$idColumn = $columns[0];
There is all the rest of it but I just need to know what on earth it is that this bit of code is expecting. I have not even got the first clue what is supposed to be sent to this thing. I just need to get it to work.
This is what I have tried so far:
function getAll($tablename)
{
$mysqlConnection = getDbConnection();//Just the normal PDO db connection
$sql = "SELECT * FROM ".$tablename;
$sth = $mysqlConnection->prepare($sql);
$sth->execute();
$resultSet = $sth->fetch(PDO::FETCH_ASSOC);
return $resultSet;
}
I have tried various different PDO::FETCH_... types but nothing is working. There is no information about what it is that I am supposed to send that part of the view.
If you want all the rows from fetch(), you will need to loop through the result set because it will return a single row. In the loop you can place them in an array.
You can use fetchAll() instead. It will return all the results as an array.
I am trying to print out the column headers for any query entered. I have other code that connects to the database and actually prints the results, but I am having trouble with the line
'$result .= $heading->name;'
I keep getting this error:
'Catchable fatal error: Object of class mysqli_result could not be converted to string in...'
Could someone explain what the problem is? According to the php manual this should give me the right information. I am fairly new to php though, so if you improve the code could you please explain how you did it?
I have the following code so far that gets a result from a query:
$result = mysqli_query($conn,$query);
if($result){
if( mysqli_num_rows($result)>0){
$columnNo = mysqli_num_fields($result);
for($i=0;$i<$columnNo;$i++){
$heading = mysqli_fetch_field_direct($result,$i);
$result .= $heading->name;
}
}
}
$result is an object being used by mysqli. Then you try to append a string onto the end of it.
Just use a different variable name besides $result which will be a string in which you will collect the names of your columns. Or you might save the names in an array like this:
$var[] = $heading->name;
since $result is an object you are using for executing the query, i won't be able to store any other data coming from db. you have to take another variable (say $variable) to store the data coming from db. Follow the below code:
$result = mysqli_query($conn,$query);
$variable=''; // add this
if($result){
if( mysqli_num_rows($result)>0){
$columnNo = mysqli_num_fields($result);
for($i=0;$i<$columnNo;$i++){
$heading = mysqli_fetch_field_direct($result,$i);
$variable .= $heading->name; //modify here
}
}
}
I hope this will solve the issue..
I'm trying to take a MySQL result row and pass it to a function for processing but the row isn't getting passed. I'm assuming this is because the actual row comes back as a object and objects can't get passed to function?
E.G
function ProcessResult($TestID,$Row){
global $ResultArray;
$ResultArray["Sub" . $TestID] = $Row["Foo"] - $Row["Bar"];
$ResultArray["Add" . $TestID] = $Row["Foo"] + $Row["Bar"];
}
$SQL = "SELECT TestID,Foo,Bar FROM TestResults WHERE TestDate !='0000-00-00 00:00:00'";
$Result= mysql_query($SQL$con);
if(!$Result){
// SQL Failed
echo "Couldn't find how many tests to get";
}else{
$nRows = mysql_num_rows($Result);
for ($i=0;$i<$nRows;$i++)
{
$Row = mysql_fetch_assoc($Result);
$TestID = $Row[TestID];
ProcessResult($TestID,$Row);
}
}
What I need is $ResultArray populated with a load of data from the MySQL query. This isn't my actual application (I know there's no need to do this for what's shown) but the principle of passing the result to a function is the same.
Is this actually possible to do some how?
Dan
mysql_query($SQL$con); should be mysql_query($SQL,$con); The first is a syntax error. Not sure if this affects your program or if it was just a typo on here.
I would recommend putting quotes around your array keys. $row[TestID] should be $row["TestID"]
The rest looks like it should work, although there are some strange ideas going on here.
Also you can do this to make your code a little cleaner.
if(!$Result){
// SQL Failed
echo "Couldn't find how many tests to get";
}else{
while($Row = mysql_fetch_assoc($Result))
{
$TestID = $Row['TestID'];
ProcessResult($TestID,$Row);
}
}
mysql_fetch_assoc() returns an associative array - see more
If you need an object, try mysql_fetch_object() function - see more
Both array and object can be passed to a function. Thus, your code seems to be correct, except for one line. It should be:
$Result= mysql_query($SQL, $con);
or just:
$Result= mysql_query($SQL);
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);
}