I've found several tutorials which have similar code like the following:
$sql = "select * from users";
$result = $conn1->Execute($sql);
if ($result==false) {
print 'error' . $conn1->ErrorMsg() . '<br>';
} else {
print_r($result->GetRows());
}
But how can $result ever be false? If I add a where clause which can not be fulfilled the else-branch is still taken since $result contains the column titles. Examples:
"select * from users"; // Select the whole table data
echo "$result";
leads to
id,username,password 1,peter,geheim 2,sabine,secret 3,thorsten,qwertz
Whereas
"select * from users where username = 'does not exist'";
echo "$result";
leads to
id,username,password
Therefore result is never false. What is my mistake here?
The Execute method returns false if the query itself fails, and not if it has 0 results.
If you want to check if the query returned any results you can use the RecordCount method.
$rows = $conn1->Execute($sql);
if ($rows->RecordCount() > 0) {
// Do something with your rows
} else {
// Nothing returned
}
Related
I have this php code:
$query = $database->query("SELECT EXISTS(SELECT * FROM contacts WHERE contact_id = '$contactID')";
if($query == 0){
echo "not registered";
}elseif($query == 1){
echo "registered"
}
If I'm not wrong, the query is suppose to return 0 or 1 and it works in my SQLite manager. What is the correct way on getting that value in Php and use it in IF ELSE statement?
If you only need a single value, you can use querySingle:
$result = $database->querySingle("SELECT EXISTS(SELECT * FROM contacts WHERE contact_id = '$contactID'");
Otherwise, with normal queries, the result returned by ->query isn't actually the data itself, but an identifier you would use to get data from the database:
$results = $db->query('SELECT bar FROM foo');
while ($row = $results->fetchArray()) {
var_dump($row);
}
Something is wrong with my php,
I'm doing an account validation where if the data exist it will display "There is data" and else "No data"...
When I enter the first 'row' reference_id and submit, it shows "There is data" which is correct but when I entered the second to the last 'row' reference_id it shows "No data" even though it exist in my Database!
Database:
reference_id (varchar 250)
status (varchar250)
PHP
if (isset($_POST['submit_valid'])) {
if (!empty($_POST['reference_id']))
{
$query = mysqli_query($con, "SELECT * FROM client_record");
$result = mysqli_fetch_array($query);
if ($result['reference_id'] == $_POST['reference_id'])
{
echo"<script type='text/javascript'> alert('There is data'); window.location.href='next_page.php'; </script>";
}
if ($result['reference_id'] !== $_POST['reference_id']) {
echo"<script type='text/javascript'> alert('No data.'); window.location.href='this_page.php'; </script>";
}
}
}
I am not sure if it's the mysqli_fetch_array fault or the if-else condition is wrong?
if you guys know the problem please help me?
Your query execution currently only looks at the first row. A fetch needs to be looped to iterate over all rows. e.g.
$query = mysqli_query($con, "SELECT * FROM client_record");
$result = mysqli_fetch_array($query);
should be
$query = mysqli_query($con, "SELECT * FROM client_record");
while($result = mysqli_fetch_array($query)) {
but this is inefficient. When looking for a specific record use a where clause. Parameterized queries also will prevent SQL injections, and quoting issues. The i in the bind_param is for an integer, if your id is a string use s.
$prepared = mysqli_prepare($con, "SELECT * FROM client_record where reference_id = ?");
mysqli_stmt_bind_param($prepared, 'i', $_POST['reference_id']);
mysqli_stmt_execute($prepared);
mysqli_stmt_store_result($prepared);
while (mysqli_stmt_fetch($prepared)) {
$query = mysqli_query($con, "SELECT * FROM client_record");
$result = mysqli_fetch_array($query);
This will give you the first row from the table.
Add a WHERE reference_id = :refid clause?!
Then bind the refid parameter, so as to avoid SQL injection.
Lapiz, the problem is actually with the comparison operator:
($result['reference_id'] == $_POST['reference_id'])
This will check the first reference_id from the returned set in array.
The best way to tackle this would be to use if (in_array(5, $result)) where 5 is the needle and $result is the array haystack.
Because all you are doing is to check if the reference exists in the returned data set .
This is also good design practices, to collect results and avoid multiple reference queries each time, hit the database once and query the result set.
If its a multidemnsional array loop through the set:
foreach($result as $resultItem)
{
if(in_array("reference_id", $resultItem, true))
{
echo "There is Data";
}
}
Good Luck .
I need to check if a record exists in a table before adding it.
I've done some digging and this is what people keep coming back too:
$result= mysql_query("SELECT id FROM mytable WHERE city = 'c7'");
if(mysql_num_rows($result) == 0) {
// row not found, do stuff...
} else {
//row found, do other stuff...
}
or some variation there of.
This logic is exactly what I need except for the fact that $result is never returning a positive result.
The record does exist and should return a positive result.
I also tried
$sql="SELECT COUNT(email) FROM table WHERE email=$mail;";
$yesorno = mysqli_query($sql);
echo $yesorno ;
as a test and the echo returns no value.
you need to check first if the query succeed running maybe there is a problem with it
$Query = "Select id from mytable where city = 'c7' ";
if($result = mysql_query($Query)) {
if ( mysql_num_rows($result) == 0 ) {
// no rows found ;
} else {
// row exist ;
}
} else {
echo "Query failed ". $Query;
}
i want to check the rows if there are any events that are binded to a host with host_id parameter, everything is well if there is not any events binded to a host, its printing out none, but if host is binded to one of the events, its not listing the events, but if i remove the codes that i pointed below with commenting problem starts here and problem ends here, it lists the events. I'm using the fetchAll function above too for another thing, there is not any such that error above there, but with the below part, it's not listing the events, how can i fix that?
Thanks
try
{
$eq = "SELECT * FROM `events` WHERE `host_id` = :id AND `confirmed` = '1' ";
$eq_check = $db->prepare($eq);
$eq_check->bindParam(':id', $id, PDO::PARAM_INT);
$eq_check->execute();
//problem starts here
$count3 = $eq_check->fetchAll();
$rowCount = count($count3);
if ($rowCount == 0)
{
echo "None";
}
//problem ends here
while($fetch = $eq_check->fetch (PDO::FETCH_ASSOC) )
{
$_loader = true;
$event_id = $fetch['event_id'];
$event_name = $fetch['event_name'];
$link = "https://www.mywebsite.com/e/$event_id";
echo "<a target=\"_blank\" href=\"$link\"><li>$event_name</li></a>";
}
}
catch(PDOException $e)
{
$log->logError($e." - ".basename(__FILE__));
}
Thank you
You can't fetch twice without executing twice as well. If you want to not just re-use your $count3 item, you can trigger closeCursor() followed by execute() again to fetch the set again.
To reuse your $count3 variable, change your while loop into: foreach($count3 as $fetch) {
The reason that it is not listing the events when you have your code is that the result set is already fetched using your fetchAll statement (The fetchAll doesn't leave anything to be fetched later with the fetch).
In this case, you might be better off running a select count(*) to get the number of rows, and then actually running your full query to loop through the results:
An example of this in PDO is here:
<?php
$sql = "SELECT COUNT(*) FROM fruit WHERE calories > 100";
if ($res = $conn->query($sql)) {
/* Check the number of rows that match the SELECT statement */
if ($res->fetchColumn() > 0) {
/* Issue the real SELECT statement and work with the results */
$sql = "SELECT name FROM fruit WHERE calories > 100";
foreach ($conn->query($sql) as $row) {
print "Name: " . $row['NAME'] . "\n";
}
}
/* No rows matched -- do something else */
else {
print "No rows matched the query.";
}
}
$res = null;
$conn = null;
?>
Note that you cannot directly use rowCount to get a count of rows selected - it is meant to show the number of rows deleted and the like instead.
I have the following code and it should return just one value (id) from mysql table. The following code doesnt work. How can I output it without creating arrays and all this stuff, just a simple output of one value.
$query = "SELECT id FROM users_entity WHERE username = 'Admin' ";
$result = map_query($query);
echo $result;
I do something like this:
<?php
$data = mysql_fetch_object($result);
echo $data->foo();
?>
You have to do some form of object creation. There's no real way around that.
You can try:
$query = "SELECT id FROM users_entity WHERE username = 'Admin' ";
//$result = map_query($query);
//echo $result;
$result = mysql_query($query); // run the query and get the result object.
if (!$result) { // check for errors.
echo 'Could not run query: ' . mysql_error();
exit;
}
$row = mysql_fetch_row($result); // get the single row.
echo $row['id']; // display the value.
all you have is a resource, you would still have to make it construct a result array if you want the output.
Check out ADO if you want to write less.
Not sure I exactly understood, what you want, but you could just do
$result = mysql_query('SELECT id FROM table WHERE area = "foo" LIMIT 1');
list($data) = mysql_fetch_assoc($result);
if you wish to execute only one row you can do like this.
$query = "SELECT id FROM users_entity WHERE username = 'Admin' ";
$result = mysql_query($query);
$row = mysql_fetch_row($result);
echo $row[0];
there have been many ways as answered above and this is just my simple example. it will echo the first row that have been executed, you can also use another option like limit clause to do the same result as answered by others above.