I'm trying to pull some information from a database, and the connection is working, but for some reason it isn't recognizing my query, even though I confirmed the query in the database with SQL and had it "generate PHP code". The echo statement is coming up blank. It's a mySQL database. Thanks for your help.
$query = "SELECT `contact` FROM `contactinfo` WHERE member=\'Henry\'";
$contact = mysqli_query($db,$query);
echo $contact;
$contact contains MySQL result object you need to fetch data from this to use this in your application.
$query = "SELECT `contact` FROM `contactinfo` WHERE member = 'Henry'";
$contact = mysqli_query($db, $query);
while ($row = mysqli_fetch_row($contact)) {
echo $row[0]; // 0 to n indicates the Column(s) Selected in SELECT Query
}
Related
I've found many similar questions regarding this but anything I try won't work.
I'm trying to run a MySQL query using the variable $epost. When I echo this variable it displays correctly, but the query returns nothing. Entering a fixed value for $epost like:
$epost='email#email.com'
Returns the correct query from the database.
$epost=mysqli_real_escape_string($conn,$_POST['email']);
echo $epost;
$sql = "SELECT memberID FROM Member WHERE email = '$epost' limit 1";
$result = mysqli_query($conn,$sql);
$row = mysqli_fetch_assoc($result);
echo $row["memberID"];
I've got two php values called $email and $pass.
email - Name of row in MySQL database
password - Name of row in MYSQL database
I'm running a sql query to select from table member where email = $email and password =$pass.
I'm then running mysqli_query to see if a row exists, I'm not getting any results. Surely the echo would echo out the ID of where the info matches.
//Get the connection info.
global $connect;
$sql = "SELECT FROM members WHERE email='$email' AND password='pass'";
//Fetch the row and store the ID of this row.
$row = mysqli_query($connect, $sql);
$id = $row['userID'];
echo $id;
Besides the fact that this code is massively exposed to SQL injection.. you are querying the data but not fetching the results.
add the fetch command:
$data = mysqli_query($connect, $sql);
$row = mysqli_fetch_assoc($data);
$id = $row['userID'];
I am having trouble selecting my data from a database and displaying it. I have looked at tutorials and i still get the same error. Some help would be appreciated. The error i am getting is couldnt fetch result.
$sql = "SELECT * FROM data";
$result = mysql_query($sql) or die("couldnt fetch result");
if($result > 0){
while ($rows = mysql_fetch_array($result)){
$username = $rows['username'];
echo $username;
}
}
Just do that (assuming got it right connecting to DB, first thing to check !)
$sql = "SELECT * FROM `data`"; // data is a reserved keyword, protect it !!!
$result = mysql_query($sql) or die("couldnt fetch result"); // potentially diying here
if($result){
while ($row = mysql_fetch_assoc($result)){
$username = $row['username'];
echo $username;
}
}
If what you're getting is literally 'couldnt fetch result' it means your mysql_query() fails, and die statement takes over. Check your database connection.
I think the very simple problem is that you check if the $result is greater then 0. But you get an resource.
$conn = mysql_connect.......
$sql = "SELECT * FROM data";
$result = mysql_query($sql) or die("couldnt fetch result");
if($result){
while ($rows = mysql_fetch_array($result)){
$username = $rows['username'];
echo $username;
}
}
And if you see your die statement you have an error in your SQL Syntax. Its very short but its possible that your table doesn't exist in that database you're trying to connect. I hope you have a connect before and its not your complete code.
You use the old mysql functions. Its better to use MySQLi or PDO.
And DATA is a reserved keyword its possible that you get problems if you use it in your query. Rename your table in prefix_data for example.
https://dev.mysql.com/doc/refman/5.7/en/keywords.html
I'm developing an app for android that uses a DB on a server.
I wrote some script php to create new rows in some tables and get all elements from a table (using JSON to exchange data between android and mysql).
Now I have a problem:
i need to select an id from a table and then use this to insert a row in anothere table that has this foreign key.
Well, when I try to select my id, i don't know why, but look like it doesn't work.
Here a simple example how I select this id:
//connect to DB...
$result = mysql_query (*SELECT id FROM 'table' WHERE name = $name );
$row = mysql_fetch_assoc($result);
$id = $row['id'];
When i use this to select an id, and put it in another query (always on the same connectio) nothing is stored.
if I force the value manually, and so in the same second query I put a number of a preesisting id, the insert works, so the problem is in this piece of code.
Hope someone could help me.
Thank you!
The code that you have put on the question, contains syntax errors.
- Remove * from the start of query
- put the query inside " "
- remove single quote ('table') from table name
Here is the modified code:
//connect to DB...
$result = mysql_query ("SELECT id FROM table WHERE name = $name" );
$row = mysql_fetch_assoc($result);
$id = $row['id'];
Also you should escape the parameter $name in query. And you should use mysqli or PDO instead of mysql extension.
try this:
$result = mysql_query (*SELECT id FROM 'table' WHERE name = $name );
$row = mysql_fetch_assoc($result);
while($row > 0){
$id = $row['id'];
}
I'm building a simple bug tracking tool.
You can create new projects, when you create a project you have to fill in a form, that form posts to project.class.php (which is this code)
$name = $_POST['name'];
$descr = $_POST['description'];
$leader = $_POST['leader'];
$email = $_POST['email'];
$sql="INSERT INTO projects (name, description, leader, email, registration_date)
VALUES ('$name', '$descr', '$leader', '$email', NOW())";
$result = mysql_real_escape_string($sql);
$result = mysql_query($sql);
if($result){
header('Location: ../projectpage.php?id='.mysql_insert_id());
}
else {
echo "There is something wrong. Try again later.";
}
mysql_close();
(It's not yet sql injection prove, far from complete...)
Eventually you get redirected to the unique project page, which is linked to the id that is stored in the MySQL db. I want to show the name of that project on the page, but it always shows the name of the first project in the database.
(here I select the data from the MySQL db.)
$query = 'SELECT CONCAT(name)
AS name FROM projects';
$result = mysql_real_escape_string($query);
$result = mysql_query ($query);
(here I show the name of the project on my page, but it's always the name of the first project in the MySQL db)
<?php
if ($row = mysql_fetch_array ($result))
echo '<h5>' . $row['name'] . '</h5>';
?>
How can I show the name of the right project? The one that is linked with the id?
Do I have the use WHERE .... ?
Yes, You have to use the WHERE to specify which project You want to get. I'm also not sure why are You using CONCAT function when You want to get only one project.
Other important thing is that You have to use mysql_real_escape_string() function on parameters before You put them in the query string. And use apropriate functions for specific type of data You receive.
So Your statement for getting the project should look like this:
SELECT name FROM projects WHERE id = ' . intval($_GET['id'])
Also when before You use the mysql_fetch_assoc() function, check if there are any records in the result with
if(mysql_num_rows($result) > 0)
{
$project = mysql_fetch_assoc($result);
/* $project['name'] */
}
try this
// first get the id, if from the url use $_GET['id']
$id = "2";
$query = "SELECT `name` FROM `projects` WHERE `id`='".intval($id). "'";
$result = mysql_query(mysql_real_escape_string($query));
use mysql_fetch_row, here you'll not have to loop through each record, just returns single row
// if you want to fetch single record from db
// then use mysql_fetch_row()
$row = mysql_fetch_row($result);
if($row) {
echo '<h5>'.$row[0].'</h5>';
}
$row[0] indicates the first field mentioned in your select query, here its name
The might be of assistance:
Your are currently assing a query string parameter projectpage.php?id=
When you access the page the sql must pick up and filter on the query string parameter like this:
$query = 'SELECT CONCAT(name) AS name FROM projects WHERE projectid ='. $_GET["id"];
$result = mysql_real_escape_string($query);
$result = mysql_query ($query);
Also maybe move mysql_insert_id() to right after assigning the result just to be safe.
$result = mysql_query($sql);
$insertId = mysql_insert_id();
Then when you assign it to the querystring just use the parameter and also the
header('Location: ../projectpage.php?id='.$insertId);