how can I get the ChatList names from the database and populate them in strings.xml file string array tag using php and mysql
<string-array name="chatListNames">
<item>Byamukama Robinhood</item>
<item>Test name</item>
<item>Ivan</item>
<item>Mohsin Afir</item>
<item>Test</item>
</string-array>
populate them in strings.xml file string array tag using php and mysql
strings.xml are for constant(read-only) strings so it's not possible to change at runtime
So instead of adding strings in strings.xml which want to change at runtime save in SharedPreferences.
how can I get the ChatList names from the database
Use web-services for getting ChatList names from server in application then update all values in SharedPreferences.
See following good tutorial for creating web-services in PHP and accessing from Android:
How to connect Android with PHP, MySQL
It is not possible in Android, you can use a database to do that. Strings.xml is can not be modified at run time.
Use the following code and change db name and access, run it in browser and copy paste the same to XML file.
<?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 ($result->num_rows > 0) {
// output data of each row
echo "<string-array name="chatListNames">";
while($row = $result->fetch_assoc()) {
echo "<item>" . $row["id"]. "</item><br>";
}
echo "</string-array>";
} else {
echo "0 results";
}
$conn->close();
?>
Related
The only tutorials I could find used unfamiliar alternative resources, like app engine, composer, github, proxies, other plugins.
Isn't it possible to simply connect to it using pure PHP?
For example, in the following code would I need to modify to get data from Google Cloud SQL?
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
?>
Google App Engine connects from inside of Google servers to the Cloud SQL instance via proxy, but the Cloud SQL instance is just a managed MySQL instance. So, if you want just to connect from the outside with PHP code and without a proxy, then you just need to authorize your IP (here: https://cloud.google.com/sql/docs/mysql/connect-external-app#appaccessIP) in the Cloud SQL instance.
And then, you should modify the code to change “localhost” to your Google Cloud SQL instance public IP (with its proper username, password and dbname). You can find the public IP here: http://console.cloud.google.com/sql/instances/
But, if you still want to have a look around PHP on Google App Engine, check that link https://cloud.google.com/appengine/docs/standard/php/cloud-sql/using-cloud-sql-mysql
I've looked all over here. Please be patient as I am new to php and mysql.
I got WAMPP installed & seems to be working OK. I created a simple "test" database from phpMyAdmin and "firsttable" in that. I can do a simple connect using example from w3schools, but trying to select & display data I entered only throws back errors.
<?php
$servername = "localhost";
$username = "root";
$password = "";
// Connect
$conn = mysqli_connect($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT reference, firstname, lastname, room FROM firsttable";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["reference"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "room:" . $row["room"]. "<br>";
}
} else {
echo "0 results";
}
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$conn->close();
?>
First off, I get a parse error on line 17. The one that reads:
if ($result->num_rows > 0) {
The error says: Trying to get property of non-object.
I tried wrapping the whole php code in tags and saving it as html, but then it appeared that no row data was ever found.
I am able to use very simple code that connects successfully. I can confirm the database is in there, so is the table, and the contents I added to it.
Please, what am I doing wrong?
You need to specify the database when you connect:
$database = 'test';
$conn = mysqli_connect($servername, $username, $password, $database);
where $database is the name of your database (test in this case). MySQL doesn't know which database your table resides in without you telling it.
In addition, you should always include error checking for your database connection (you have two of these, you don't need the last one) as well as any queries. Sans this, you can check your error logs for more information when something fails.
I am new in PHP and would need some explanation. Here is a code where we connect to MySQL with PHP. Can you please explain me where is the statement that makes the connection? I can see only that we define what the value of $conn is, but does it mean execution as well? The other thing is: where do we create the database? I can see that we give the string "CREATE DATABASE myDB" as a value to $sql and we have an if statement, but does the expression ($conn->query($sql) === TRUE) also evaluated? It is strange for me, can somebody please explain it to me?! :) Thanks!
<?php
$servername = "localhost";
$username = "username";
$password = "password";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// Create database
$sql = "CREATE DATABASE myDB";
if ($conn->query($sql) === TRUE) {
echo "Database created successfully";
} else {
echo "Error creating database: " . $conn->error;
}
$conn->close();
?>
Here is a simple explanation of which lines do what. If you would like to know specifically what the individual parts of these mean, then please say which ones so they can be further explained to you. Or the correct links pointed to.
I notice that you are using the W3Schools example, as an almost exact copy and paste. Have you installed MySQL on your machine and created a username and password?
<?php
$servername = "localhost"; // This is the location of your server running MySQL
$username = "username"; // This is the username for MySQL
$password = "password"; // This is the password for MySQL
// Create connection
$conn = new mysqli($servername, $username, $password); // This is where you create a connection
// Check connection
if ($conn->connect_error) { // This checks if the connection happened
die("Connection failed: " . $conn->connect_error); // and produces an error message if not
} // otherwise we move on
// Create database
$sql = "CREATE DATABASE myDB"; // This is the SQL query which is sent to the MySQL server
if ($conn->query($sql) === TRUE) { // When the if statement begins here, it executes the query and test if it returns true
echo "Database created successfully"; // If it returns true then here is the message is returns
}
else {
echo "Error creating database: " . $conn->error; // Or if there was error with the query this is returned
}
$conn->close(); // Close the connection when it is no longer in use
?>
Although, your question does not belong here (This place is to help with your coding issues), but I will give you a bit explanation.
PHP reads each line and EXECUTES It. the create connection part opens a new connection using the "new" object and save it a variable ($conn),
($conn->connect_error) checks if the connection was successful with connect_error property. if it was connected, continue, or else through and error and stop.
If connection was successful, then create the database based on connection opened in variable ($conn).
I have created one HTML form which takes input from user ,Now I need to search user inputed name in Mysql database and print details related to that user inputed name which is stored in Mysql database.
Below script is creating HTML form to take user input, Saved as "ProcessTracking.html".
<form action="details.php" method="get"/>
<h3 align="center"><FONT color=#CCFF66>ENTER SO NUMBER</h3>
<p align="center">
<input type="text" id="SO_Number" name="SO_Number"/>
</p>
<div style="text-align:center">
<button type="submit" value="SEARCH">
<img alt="ok" src=
"http://www.blueprintcss.org/blueprint/plugins/buttons/icons/tick.png"/>
SEARCH
</button>
</form>
Below PHP script named as "details.php"
<?php
$userinput = $_GET['SO_Number'];
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ProcessTrackingSystem";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_errno) {
printf("Connect failed: %s\n", $conn->connect_error);
exit();
}
$result = mysqli_query($conn, "SELECT * FROM ProcessTrackingSystem.ProcessDetails WHERE SO_Number = '$userinput'") or die(mysqli_error($conn));
$row = mysqli_fetch_assoc($result);
#printf ("SO_Number: %s \n",$row["SO_Number"])
#print_r($row);
printf ("SO_Number:");
printf($row["SO_Number"]);
printf ('--||--');
printf ("Name:");
printf($row["Name"]);
printf ('--||--');
$conn->close();
?>
Firstlly you are not using the $_GET['SO_Number'] parameter in a WHERE of SQL statement. Secondlly you are using both mysql and mysqli which are totaly diffrent and don't work together. For usage see mysqli_fetch_row() and mysqli_query(). Also use print_r($row);.
Here is the corrected code:
<?php
$userinput = $_GET['SO_Number'];
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ProcessTrackingSystem";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_errno) {
printf("Connect failed: %s\n", $conn->connect_error);
exit();
}
$result = mysqli_query($conn, "SELECT * FROM ProcessTrackingSystem.ProcessDetails WHERE SO_Number = '$userinput'") or die(mysqli_error($conn));
$row = mysqli_fetch_row($result);
print_r($row);
$conn->close();
?>
EDIT: Added code example.
You have mixed mysql and mysqli api together. Try using either one.
Note: mysql api is deprectaed as of php 5.5.0
1st Error
As saty says it is because of the syntax error you have in this line
print_r"$row";
which should be as print_r($row)
2nd Error
You're mixing mysql & mysqli
I am not sure about the table that you have.
Recommendation :
I would recommend you to turn on the error_reporting if not those errors will be in your errors_log file
Also for debugging your sql, you can first construct your sql query, run in the phpmyadmin or related tools for your query, then fire the query and make this done.
Note :
If you are using these code in online then the error_log will be in the directory where you execute this page. (But it may change according to your hosting)
If you are running in local machine the error log may locate according to the server you use...
You can find by printing the php's configuration by phpinfo and find for
error_log
May this thing will fix your issue
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ProcessTrackingSystem";
$so = $_POST['SO_Number'];
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$stmt = $conn->prepare("SELECT * FROM ProcessDetails WHERE SO_Number=$so");
$stmt->execute();
$result = $stmt->get_result();
$row = $result->fetch_assoc();
print_r($row[SO_Number]);
$conn->close();
?>
I'm an iOS developer trying to update a bit 'o php to work with the mysqli_* stuff instead of the deprecated mysql_* stuff.
I know next to nothing about php/mysql, and I'm stuck. I'm using a script found at http://www.w3schools.com/php/php_mysql_select.asp, with changes to reflect my field names etc
When I call the following from a browser I get an access error (Connection failed: Access denied for user 'whoisit7'#'localhost' (using password: YES)). The server name, password and username etc I'm using all work with an existing script in a browser but not with this new one.
Can anyone point me at what I'm getting wrong?
<?php
$servername = "localhost";
$userName = "whoisit7_ios";
$password = "blahblahblah";
$dbName = "whoisit7_scathing";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT name, latitude, longitude FROM location where status = 1";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "name: " . $row["name"]. " - latlong: " . $row["latitude"]. " " . $row["longitude"]. "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
?>
In PHP, variable names are case-sensitive. So, Change your connection variable names to:
$servername = "localhost";
$username = "whoisit7_ios";
$password = "blahblahblah";
$dbname = "whoisit7_scathing";
Read it if you are new to PHP: http://php.net/manual/en/language.variables.basics.php
Php is case sensitive, so change the below line from:
$conn = mysqli_connect($servername, $username, $password, $dbname);
to
$conn = mysqli_connect($servername, $userName, $password, $dbName);
You have made error in $username which is $userName and $dbname which should be $dbName as you declared above.
Just change -->
localhost to 127.0.0.1 //you need to change params in mysql config if u want to use it as localhost
And
$username to $userName //As u need to keep these variables similar :)
this Problem seems to be coming from database privileges as i have also have encountered this before. Give this fix a try.
GRANT ALL PRIVILEGES ON database_name TO usernamehere#host IDENTIFIED BY 'yourpassword';
FLUSH PRIVILEGES;
Just change the lower case letters to your own credentials.