show echo if column is empty else show another echo - php

i wrote this code but there is some mistake its only shows the first echo even if the column is not empty.
i wnat the code to show echo "1" if the column subuser1 is empty
else to show echo "2" if the column subuser1 is not empty.
<?php
include_once 'dbconnect.php';
// Create connection
$conn = new mysqli(DBHOST,DBUSER,DBPASS,DBNAME);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql= ("SELECT subuser1 FROM users WHERE id=".$_SESSION['user'] );
$result = $conn->query($sql);
$subuser1 = $row["subuser1"];
if ($result->num_rows > 0) {
if (empty($subuser1)) {
echo "subuser1 is either 0, empty, or not set at all";
}
} else {
echo " subuser1 not empty";

I solve it like that it works perfect if anyone need it.
<?php
$userR= 0 ;
include_once 'dbconnect.php';
// Create connection
$conn = new mysqli(DBHOST,DBUSER,DBPASS,DBNAME);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql= ("SELECT * FROM users WHERE id=".$_SESSION['user'] );
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
if ($row["subuser1"] == NULL ){
echo ""; }
else {
echo " bla bla bla "
}
}
?>

Related

Error when trying to display my array?

So I am trying to display an array from a database that I have. When I run the script the script, I get an internal server error. Now I am not sure if this has to do with my config script or if I am not cycling through my array properly.
include 'config.php';
$conn = name2;
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Feild FROM Season 1";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "Feild " . $row["Feild"]. " "<br>";
}
} else {
echo "0 results";
}
The syntax you have used to display the results was incorrect, replace:
echo "Feild " . $row["Feild"]. " "<br>";
With:
echo 'Feild '.$row["Feild"].'<br>';
You have not terminated your string properly.
Replace
echo "Feild " . $row["Feild"]. " "<br>";
With
echo "Feild " . $row["Feild"]. "<br>";
If you had errors turned on you would be getting an error stating a line number.
We need to know which column you're selecting from, as "SELECT Feild FROM Season 1" isn't valid. It should be something like "SELECT Feild FROM Season WHERE column = '1'"
With that in mind, this gets you closer to a solution:
include 'config.php';
$conn = name2;
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Feild FROM Season 1";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "Feild ". $row["Feild"] ."<br>";
}
} else {
echo "0 results";
}
Use correct syntax
echo "Feild ".$row['Feild']."<br>";

Is it possible to assign sql table value to a variable in php?

Basically what I want is to assign the value of a field in my database to an variable. Can it be done in an effective way?
I was thinking something like:
$sql2 = mysql_query("SELECT * FROM rom WHERE idrom = 101");
while ($row = mysql_fetch_array($sql2)) {
$rom1 = $row['idrom'];
$status = $row['status'];
echo $rom1;
echo $status;
}
But this doesn't echo anything.
Edit:
I have gotten a bit longer on the way, now I am looking for a simpler way to assign the values to variables. As we speak I only need 4 values, but this still doesn't look like a very good way to accomplish what I want. Any better suggestions?
Heres what I got now:
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM rom WHERE idrom = 101";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$room101 = $row["idrom"];
$status101 = $row["status"];
echo "This is roomnumber ". $room101 . "!<br >";
echo "And the status of roomnumber ". $room101 ." is ". $status101 ."<br><br>";
}
} else {
echo "0 results";
}
$sql2 = "SELECT * FROM rom WHERE idrom = 102";
$result2 = $conn->query($sql2);
if ($result2->num_rows > 0) {
// output data of each row
while($row = $result2->fetch_assoc()) {
$room102 = $row["idrom"];
$status102 = $row["status"];
echo "This is roomnumber ". $room102 . "!<br >";
echo "And the status of roomnumber ". $room102 ." is ". $status102 ."<br><br>";
}
} else {
echo "0 results";
}
You can use bind_result to do that.
Please try this simple example, hope it run well :
$mysqli = mysqli_connect('host', 'user', 'pass','dbase')or die('Could not connect: ' . mysqli_error());
if ($mysqli->connect_error) {
die("Connection failed: " . $mysqli->connect_error);
}
// Because you provide an Id in where clause, you can use it for the new variable
$output = array();
$id = 101;
$sql = "select idrom,status from rom where idrom=?";
$stmt = $mysqli->prepare($sql);
$stmt->bind_param('i',$id);
$stmt->execute();
if ($stmt->errno == 0) {
$stmt->store_result();
// $stmt->bind_result($idrom,$status); // way 1
$stmt->bind_result($output[$id],$output["status".$id]); // way 2
while ($stmt->fetch()) {
echo $output[$id]." -> ".$output["status".$id]."<br />"; // So, your output variable will be like $output[101] for way 2
// or you defined it here like :
// $output[$idrom] = $idrom;
// $output["status".$idrom] = $status; // For way 1
}
} else {
return "Error: " . $sql . "<br>" . $stmt->error;
}
$stmt->close();
$mysqli->close();

CONCAT function in MySQL using PHP Undefined Variable

I need help on how to do this correctly. I need to execute this command:
SELECT concat(branchname, -->, itemtype, '(, quantity, ')') from monitoring
order by itemtype;
the syntax works in MySQL console. However, im having trouble with implementing it on php. I always get
"Undefined index: branchname"
"Undefined index: itemtype"
"Undefined index: quantity"
using this code:
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "dex_test";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT concat(branchname,itemtype,quantity) from monitoring order by itemtype";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo " " . $row["branchname"]. " " . $row["itemtype"]. " ".$row["quantity"]. "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
The error says it's in this line
echo " " . $row["branchname"]. " " . $row["itemtype"]. " ".$row["quantity"]. "<br>";
Im confused because I basically ran the same code that worked that lets me see the itemtype in the table:
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "dex_test";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT itemtype FROM monitoring";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "itemtype: " . $row["itemtype"]. "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
Help anyone?
It seems your query needs update
"SELECT concat(branchname,itemtype,quantity) from monitoring order by itemtype";
It should be
"SELECT branchname,itemtype,quantity from monitoring order by itemtype";
I have posted this answer in reference of how you were calling your fields in while loop
echo " " . $row["branchname"]. " " . $row["itemtype"]. " ".$row["quantity"]. "<br>";
and if you need to show the concat value within one field than it should be something like
$sql = "SELECT concat(branchname,' ',itemtype,' ',quantity) as branch from monitoring order by itemtype";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo $row["branch"]."<br>";
}
} else {
echo "0 results";
}
Just define the alias for the concatenated columns. Use this -
SELECT concat(branchname,itemtype,quantity) as branchname from monitoring order by itemtype
Or if you want them seperately then -
SELECT branchname, itemtype, quantityfrom monitoring order by itemtype

If 0 items in DB table, return error

I am looking for a way to display an error message if there is nothing listed in the table.
I have a photos table.
If this tables is empty, id like to echo something.
else, show the pictures.
inside of that table I have
id, name, url
id = id
name = name of image
url = url of image.
If there are no rows, we have an error.
$query1 = mysql_query("SELECT COUNT(*) FROM photos;");
mysql_fetch_array($query1);
if(empty($query1)) {
echo "nothing";
} else {
echo "good";
}
Try this,
$query = "SELECT * FROM photos";
$result= mysql_query($query);
$length= mysql_num_rows($result);
if($length>0)
{
while($rows = mysql_fetch_array($result))
{
echo $rows['name'];
echo "<img src='$rows[url]' />";
}
}
else
{
echo "Nothing to display";
}
Hope this will work
What about something like...
$sql = "SELECT COUNT(*) AS amountPhotos FROM photos";
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
if ($row["amountPhotos"] == 0) {
echo "There are no photos in the photo table.";
}
or
$sql = "SELECT * FROM photos LIMIT 1";
$result = mysql_query($sql);
if (mysql_num_rows($result) == 0) {
echo "There are no photos in the photo table.";
}
Try this
$query1 = mysql_query("SELECT COUNT(*) FROM photos;");
$result = mysql_fetch_array($query1);
if(empty($result)) {
echo "nothing";
} else {
echo "good";
}
This pretty much sums up the answer for this question: http://www.w3schools.com/php/php_mysql_select.asp
They even provided a sample code:
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = $conn->query($sql);
if ($result->num_rows > 0) { //<--- here they check if number of rows returned is greater than 0 (so there is data to display)
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>";
}
} else {
echo "0 results"; //<----- nothing found
}
$conn->close();
?>
Just modify this and you'll be good to go.

check if row exists else create one

Motive: I just want to crreate a row tih specified data if one with the same data does not exists.
Thing I have tried so far is -
CODE:
<?php
if(empty($_GET['a'])) {$xyz ="new";} else{$xyz=$_GET['a'];}
$servername = "mysql.soemwhere.com";
$username = "u130204422_acb";
$password = "YES-I-KNOW";
$dbname = "u130204422_acb";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$output ="SELECT * FROM trailers
WHERE url='$xyz' LIMIT 1";
$result = mysqli_query($conn,$output);
while($row = mysqli_fetch_array($result)) {
$pid=$row["title"];
echo $pid;}
if (mysqli_num_rows($result) > 0) {echo 'yes';}
else{
$msql = "INSERT INTO `trailers`(`url`,`title`) VALUES ('$xyz','dekhlia')";
if ($conn->query($msql) === TRUE) {
echo "New record created successfully. Refrsh the page and it will echo dekhlia";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;}
echo "hehe";
$conn->close();}
?>
I am unable to figure out why this code is'nt working.
You are unnecessarily adding concatenation in the SQL string.
Its a part of the sting, so, remove it.
Change
$output ="SELECT * FROM trailers
WHERE url='.$xyz.' LIMIT 1";
To:
$output ="SELECT * FROM trailers
WHERE url='$xyz' LIMIT 1";

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