I am trying to display or return the Date value from a selected user
function mysql_select_one_data($database, $table, $data, $country, $phone){
$conn = mysql_connect("localhost", "root", "test123");
mysql_select_db($database, $conn) or die (mysql_error());
$query = "SELECT DATE FROM Temp_Users WHERE Country = '+94' AND Phone = '1234539543';";
$result = mysql_query($query) or die (mysql_error());
if (mysql_num_rows($result) > 0) {
// output data of each row
while($row = mysql_fetch_assoc($result)) {
echo "Date: " . $row["Date"]. "<br>";
}
} else {
echo "0 results";
}
return $result;
}
result:
Date:
Resource id #5
but when I do the query in phpmyadmin it returns
I am new in Php and I Don't know what I am doing wrong, Thanks for your help.
change this
$query = "SELECT DATE FROM Temp_Users WHERE Country = \"+94\" AND Phone = \"1234539543\";";
to this
$query = "SELECT DATE FROM Temp_Users WHERE Country = '+94' AND Phone = '1234539543';";
Change the echo command to echo "Date: $row['DATE']<br>"
Related
I'm trying to make a Check-in/out system.
So far I have a dropdown that get the list of active events.
<select name="events">
<?php
$conn = new mysqli('localhost:3306', 'user', 'pw', 'database') or die ('Cannot connect to db');
$eveny = $conn->query("select event_title from events_event where inactive=0");
while ($row=mysqli_fetch_array($eveny)) {
unset($event);
$event = $row['event_title'];
echo '<option value="'.$event.'">'.$event.'</option>';
}
?>
</select>
And a textbox that searches users based on first name, but it auto displays results (like a Google search) and then fills out the info with both First Name and Last name. Source.
The only change in the php is the echo to show both first and last names as follows:
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
echo "<p>" . $row["per_FirstName"] . " " . $row["per_LastName"] . "</p>";
}
NOW FOR THE PROBLEM
I have made the frontend into a form, and a submit button using method="post".
But something in my php is not functioning/lacking.
<?php
$db = new mysqli('localhost:3306', 'user', 'pw', 'database') or die ('Cannot connect to db');
session_start();
if($_SERVER["REQUEST_METHOD"] == "POST") {
$myname = mysqli_real_escape_string($db,$_POST['fullname']);
$eventy = mysqli_real_escape_string($db,$_POST['events']);
//$checktime = mysqli_real_escape_string($db,date("Y-m-d H:i:s"));
$evid = "SELECT event_id from events_event where event_title = '$eventy'";
$revvy = mysqli_query($db,$evid);
$nameParts = explode(' ', $myname);
$firstName = trim($nameParts[0]);
$lastName = trim($nameParts[1]);
$sql = "SELECT per_ID FROM person_per WHERE per_FirstName = '$firstName' AND per_LastName = '$lastName'";
$result = mysqli_query($db,$sql);
//$row = mysqli_fetch_row($result);
//$result = $result->fetch_all();
while($row = mysqli_fetch_assoc($result)){
$perID = $row['per_ID'];
}
while($row2 = mysqli_fetch_assoc($revvy)){
$evvy = $row['event_ID'];
}
$count = mysqli_num_rows($result);
// table row must be 1 row if it succeeded
if($count == 1) {
//session_register("myname");
//$_SESSION['login_user'] = $myname;
$checkin = "insert into event_attend (attend_id, event_id, person_id, checkin_date) values (DEFAULT, '$evvy', '$perID', now())" or die(mysqli_error());;
mysqli_query($db,$checkin);
header("location: checkedin.php");
}else {
$error = "An error occurred.";
}
}
?>
The $myname, is the result of both first name and last name, I need just First Name based on the filled out text field which uses both first and last names.
I also can't get the Event_ID from the dropdown.
If user's first and last name are separated by space:
$nameParts = explode(' ', $myname);
$firstName = trim($nameParts[0]);
how would i get this to display a message in place of the Query if no result is found i updated the code but its just showing "N"
<?php
$hostname = "...";
$username = "";
$password = "";
$db = "";
$dbconnect=mysqli_connect($hostname,$username,$password,$db);
if ($dbconnect->connect_error) {
die("Database connection failed: " . $dbconnect->connect_error);
}
$query=mysqli_query($dbconnect,"SELECT DISTINCT companyname,client_id,feedback,status from review WHERE status=1 ORDER BY RAND() LIMIT 4");
$rows_get = mysqli_num_rows($query);
if ($rows_get >0)
{
$query2=mysqli_query($dbconnect,"SELECT DISTINCT companyname,client_id,feedback,status from review WHERE status=1 ORDER BY RAND() LIMIT 4");
$row1 = mysqli_fetch_assoc($query2);
$row2 = mysqli_fetch_assoc($query2);
$row3 = mysqli_fetch_assoc($query2);
$row4 = mysqli_fetch_assoc($query2);
$row5 = mysqli_fetch_assoc($query2);
}else {
$row1 = "N0 Data";
$row2 = "N0 Data";
$row3 = "N0 Data";
$row4 = "N0 Data";
$row5 = "N0 Data";
}
?>
Do as follows:
After $query insert this:
$rows_get = mysqli_num_rows($query);
if ($rows_get >0)
{
//do all database operation
}else {
echo " No data found";
}
Hope this helps.
Amend your code for example..
if ($row_get>0){
//i assume you are getting multiple rows
while ($data =mysqli_fetch_assoc ($query))
{
//run this loop and you will get all you rows.
}
}
I have over 40'000 entries and each is assigned to a "list_name"
I am basically trying to get just the list_name value echo'd out
$groupq = mysqli_query($dbc, "SELECT * FROM `products-full` GROUP BY `list_name`");
$groupr = mysqli_fetch_assoc($groupq);
do {
echo $groupr['list_name'];
} while($groupr = mysqli_fetch_assoc($groupq));
however its only displaying 1 entry then no more ..
https://imgur.com/a/3rnXGet
Try this.
$groupq = mysqli_query($dbc, "SELECT * FROM `products-full` GROUP BY `list_name`");
while($groupr = mysqli_fetch_assoc($groupq)){
echo $groupr['list_name'];
}
$database = "sample" //replace your database name here
$conn=new mysqli("localhost","root","",$database); // here username is root and password is null , change it according to yours
if($conn->connect_error)
{
echo $conn->connect_error;
die("sorry database connection failed");
}
$sql = "SELECT * FROM products-full GROUP BY list_name";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo $row['list_name'];
}
}
thats it
try this
$groupq = mysqli_query($dbc, "SELECT list_name, count(*) FROM `products-full` GROUP BY
`list_name`");
while($groupr = mysqli_fetch_assoc($groupq)) {
echo $groupr['list_name'];
}
I am trying to update a column of a database's table using this
$url = mysqli_connect($servername, $dbusername, $usrpassword, $dbname);
// Check connection
if (!$url) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT adv_val FROM current_advert";
if($result = mysqli_query($url, $sql)){
while($row = mysqli_fetch_assoc($result)){
$adv_val = $row['adv_val'];
}
}
echo "current advert is:" . $adv_val;
// Attempt select query execution
$sql = "SELECT advert_text FROM advertisements WHERE advert_id = $adv_val";
if($result = mysqli_query($url, $sql)){
while($row = mysqli_fetch_assoc($result)){
$advert = $row['advert_text'];
}
}
echo "<br>current advert text is:" . $advert;
if ($adv_val == 1 OR $adv_val == 2 OR $adv_val == 3) {
$adv_val = $adv_val + 1;
} else {
$adv_val = 1;
}
$sql = "UPDATE current_advert SET adv_val='$adv_val'";
// Close connection
echo "<br>next advert id is:" . $adv_val;
mysqli_close($url);
The connection to the database is ok since i'm able to read data from it in the beginning of my script. This is killing me!
$sql = "UPDATE current_advert SET adv_val='$adv_val'";
mysqli_query($url, $sql);
you are missing the 2nd line.
I am trying to draw data from multiple tables that have been indexed to relate to one another. I ran this query in MySQLWorkbench, and it ran successfully. However when I tried to run a PHP test, nothing showed up, not even for the first field. Here is my code:
<?php
$db = new mysqli('host', 'user', 'password', 'database');
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$query = "
SELECT
`Contact`.`firstName`,
`Contact`.`lastName`,
`ssn`.`ssn`,
`Contact`.`country`,
`Allergies`.`allergy`,
`Allergies`.`allergyType`,
`Allergies_Contact`.`allergyNotes`,
`CurrentPrescriptions`.`prescriptionName`,
`CurrentPrescriptions`.`prescribedDate`,
`BloodType`.`bloodType`
FROM
`database`.`Contact`,
`database`.`Allergies_Contact`,
`database`.`Allergies`,
`database`.`ssn`,
`database`.`CurrentPrescriptions`,
`database`.`BloodType`
WHERE
`Contact`.`contactKey` = `Allergies_Contact`.`contactKey`
AND `Allergies`.`allergiesKey` = `Allergies_Contact`.`allergiesKey`
AND `ssn`.`contactKey` = `Contact`.`contactKey`
AND `CurrentPrescriptions`.`contactKey` = `Contact`.`contactKey`
AND `BloodType`.`contactKey` = `Contact`.`contactKey`;
";
$result = $db->query($query) or die($db->error.__LINE__);
if ($result = mysqli_query($db, $query)) {
while ($row = mysqli_fetch_row($result)) {
print(row[0]);
}
mysqli_free_result($result);
}
mysqli_close($db);
?>
Please tell me what I am doing wrong here, because from what I can see its formatted correctly.
Several things:
1.- You have two query sentences, change:
$result = $db->query($query) or die($db->error.__LINE__);
if ($result = mysqli_query($db, $query)) {
With this
$result = $db->query($query) or die($db->error.__LINE__);
if ($result !== false) {
2.- Yo made a mistake when trying to print the variable, change:
while ($row = mysqli_fetch_row($result)) {
print(row[0]);
}
With this
while ($row = mysqli_fetch_row($result)) {
print($row[0]); // You missed a $
}
<?php
//conection:
$link = mysqli_connect("myhost","myuser","mypassw","mybd") or die("Error " . mysqli_error($link));
//consultation:
$query = "SELECT name FROM mytable" or die("Error in the consult.." . mysqli_error($link));
//execute the query.
$result = $link->query($query);
//display information:
while($row = mysqli_fetch_array($result)) {
echo $row["name"] . "<br>";
}
?>
http://php.net/manual/en/function.mysqli-connect.php