if ($con->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM Accounts WHERE acccount_num ='$accountNum'";
$result = $con->query($sql);
//$row = $result->fetch_row();
$row = mysqli_fetch_assoc($result);
print (json_encode($row));
$con->close();
This is my php code for connecting to the database where it is failing. I checked the field names in the actual database and I also checked all my code in the java project where my code is actually using the database. Really new to the whole concept of databases and PHP.
Use this
$sql = "SELECT * FROM Accounts WHERE acccount_num =$accountNum";
instead of
$sql = "SELECT * FROM Accounts WHERE acccount_num ='$accountNum'";
Try this or maybe you have an extra c in acccount_num.
$sql = "SELECT * FROM `accounts` WHERE `acccount_num` ='".$accountNum."'"
Related
I'm just trying to get something basic to appear for now and it's not working. It should display 1 on the screen. Is my logic wrong? I paste my statement in console and get 1.
<?php
$sql = mysqli_query("Select moist_measure_avail from sigh_in_account where moist_measure_avail = '1'");
$result = mysqli_fetch_array($sql);
echo $result['moist_measure_avail'];
First of all you should have to create a connection
<?php
$conn = mysqli_connect("localhost "," root","","dbname");
?>
And then you have to include the conn variable in your query
$sql = mysqli_query( $conn, " SELECT moist_measure_avail from sigh_in_account WHERE moist_measure_avail = '1'");
The sql statements must be either all caps or all small
$sql = mysqli_query("Select moist_measure_avail from sigh_in_account where moist_measure_avail = '1'");
pass connection as first param in above method. something like this
$sql = mysqli_query($conn,"Select moist_measure_avail from sigh_in_account where moist_measure_avail = '1'");
where $conn is mysql connection
I want to execute a query that i saved in my database like this:
ID | NAME | QUERY
1 | show_names | "SELECT names.first, names.last FROM names;"
2 | show_5_cities | "SELECT cities.city FROM city WHERE id = 4;"
Is this possible ?
I am kinda noob in php so plz explain if it is possible.
If I understand you correctly, you have your queries saved in the database in a table and you want to execute those.
Break the problem down: you have two tasks to do:
Query the database for the query you want to run.
Execute that query.
It's a bit meta, but meh :)
WARNING: the mysql_ functions in PHP are deprecated and can be dangerous in the wrong hands.
<?php
if (!$link = mysql_connect('mysql_host', 'mysql_user', 'mysql_password')) {
die('Could not connect to mysql');
}
if (!mysql_select_db('mysql_dbname', $link)) {
die('Could not select database');
}
$name = "show_5_cities"; // or get the name from somewhere, e.g. $_GET.
$name = mysql_real_escape_string($name); // sanitize, this is important!
$sql = "SELECT `query` FROM `queries` WHERE `name` = '$name'"; // I should be using parameters here...
$result = mysql_query($sql, $link);
if (!$result) {
die("DB Error, could not query the database\n" . mysql_error(););
}
$query2 = mysql_fetch_array($result);
// Improving the code here is an exercise for the reader.
$result = mysql_query($query2[0]);
?>
if you did create a stored procedure/function you can simply use:
mysql_query("Call procedure_name(#params)")
Thats will work. reference here: http://php.net/manual/en/mysqli.quickstart.stored-procedures.php
Querying the table to get the query, then executing that query and looping through the results and outputting the fields
<?php
$link = mysqli_connect("localhost", "my_user", "my_password", "world");
/* check connection */
if (mysqli_connect_errno())
{
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$RequiredQuery = intval($_REQUEST['RequiredQuery']);
$sql = "SELECT `QUERY` FROM QueryTable WHERE ID = $RequiredQuery";
$result = mysqli_query($link, $sql);
if ($row = mysqli_fetch_assoc($result))
{
$sql = "SELECT `QUERY` FROM QueryTable WHERE ID = $RequiredQuery";
$result = mysqli_query($link, $row['QUERY']);
while ($row2 = mysqli_fetch_assoc($result))
{
foreach($row2 AS $aField=>$aValue)
{
echo "$aField \t $aValue \r\n";
}
}
}
?>
just open the Table and get the individual query in a variable like
$data = mysql_query('SELECT * FROM <the Table that contains your Queries>');
while(($row = mysql_fetch_row($data)) != NULL)
{
$query = $row['Query'];
mysql_query($query); // The Query from the Table will be Executed Individually in a loop
}
if you want to execute a single query from the table, you have to select the query using WHERE Clause.
I'm trying to count entries in a database based on 2 basic criteria. It is returning a blank result, even though there are results to be found. Anyone have any idea what I am doing wrong here? I have tried it so many different ways and they all return no result. (If I enter the query directly in phpmyadmin it returns a result.)
$sql = "SELECT count(*) as total_count from orderOption3Detail WHERE orderDate='$orderDate' AND studentID='$studentID'";
$numericalResult = mysql_query($sql, $con);
$row = mysql_fetch_object($numericalResult);
$totalOrders1 = $row->total_count;
echo "My orders:" . $totalOrders1;
As others stated, make sure you sanitize variables before they go into query.
$sql = "SELECT * FROM orderOption3Detail WHERE orderDate = '" . $orderDate . "' AND studentID = '" . $studentID . "'";
$sql_request_data = mysql_query($sql) or die(mysql_error());
$sql_request_data_count = mysql_num_rows($sql_request_data);
echo "Number of rows found: " . $sql_request_data_count;
That's all you need.
Edited: providing full code corrected:
$con=mysqli_connect($db_host,$db_user,$db_pass,$db_name); // Check connection
if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } //global option 1
$sql = "SELECT count(*) as total_count from orderOption3Detail WHERE orderDate='$orderDate' AND studentID='$studentID'";
//echo $sql;
$numericalResult = $con->query($sql);
$row = mysqli_fetch_object($numericalResult);
echo $row->total_count; //echo (int) $row->total_count;
Please test this and let me know. Good luck!
----- End Editing ----
Have you tested assigning values directly as a test in your SQL string, like:
$sql = "SELECT count(*) as total_count from orderOption3Detail WHERE orderDate='05/23/2012' AND studentID='17'";
Also, did you check if the date's format is correct, reading that $orderdate variable and testing it in PHPMyAdmin?
Did you read the $sql with values inserted and test in PHPMyAdmin and worked?
Also, check the connection to assure there is no problem there.
One more thing, sorry. You seem to be using the wrong syntax in your mysql_query statement. That way works for mysqli_query, and the parameters would be inverted. Try only:
$numericalResult = mysql_query($sql);
Provided you made the connection and database selection previously, like in:
$connection=mysql_connect($db_host, $db_username, $db_password);
if (!$connection)
{
$result=FALSE;
die('Error connecting to database: ' . mysql_error());
}
// Selects database
mysql_select_db($db_database, $connection);
Best wishes,
I am using php to get records from a mysql database using the following code:
<?php
$username="";
$password="";
$database="";
$hostname="";
$con = mysql_connect($hostname, $username, $password);
if (!$con){
die('Could not connect: ' . mysql_error());
}
mysql_select_db($database, $con);
if(isset($_POST['emp'])){
$emp = $_POST['emp'];
$result = mysql_query("SELECT * FROM contact_log", $con);
echo mysql_num_rows($result);
die();
while($row = mysql_fetch_array($result)){
$emp = $row['emp'];
echo $emp.'<br>';
}
die();
}
mysql_close($con);
?>
This works fine and returns the correct fields. The problem is that if I change the query to
$result = mysql_query("SELECT DISTINCT * FROM contact_log", $con);
or
$result = mysql_query("SELECT * FROM contact_log GROUP BY emp", $con);
no results are returned.
mysql_num_rows does not even return a value which indicates to me that those lines are breaking my code but I am unable to figure out how.
I doubt you want to do a distinct * on your first query. Looking at your code, you probably want:
"SELECT DISTINCT emp FROM contact_log"
And you can get more information about what is going wrong with mysql_error:
mysql_query("select * from table") or die(mysql_error())
Finally, are you sure that $_POST['emp'] is being sent? Put an echo right after that if to make sure. And just so you know, you aren't using the emp POST variable for anything other than a flag to enter that block of code. $emp = $_POST['emp']; is doing absolutely nothing.
I have read multiple question in the internet including this stackoverflow question but none of them working for me. Here is my code:
<?php
$conn1 = mysql_connect("localhost","root","passw0rd") or die(mysql_error());
$conn2 = mysql_connect("localhost","root","passw0rd") or die(mysql_error());
mysql_select_db("asteriskcdrdb",$conn1);
mysql_select_db("pj8v2",$conn2);
$query = "SELECT * FROM cdr";
$result = mysql_query($query,$conn1);
var_dump($result);
$query2 = "SELECT * FROM tb_did_avalaible";
$result2 = mysql_query($query2,$conn2);
var_dump($result2);
?>
When i var_dump the result, it return false. What is the problem here? Thank you.
You dont need two connections, if both databases are located on the same mysql-server and you access them both as unique user.
You also don't need to select a DB.
Just use the database-name as prefix when specifying the tables:
<?php
mysql_connect("localhost","root","pass") or die(mysql_error());
$query = "SELECT * FROM asteriskcdrdb.cdr";
$result = mysql_query($query)or die(mysql_error());
var_dump($result);
$query2 = "SELECT * FROM pj8v2.tb_did_avalaible";
$result2 = mysql_query($query2)or die(mysql_error());
var_dump($result2);
?>
The real problem in your code is: there can only be one active DB, it should work this way:
<?php
$conn1 = mysql_connect("localhost","root","passw0rd") or die(mysql_error());
$conn2 = mysql_connect("localhost","root","passw0rd",true) or die(mysql_error());
mysql_select_db("asteriskcdrdb",$conn1);
$query = "SELECT * FROM cdr";
$result = mysql_query($query,$conn1);
var_dump($result);
mysql_select_db("pj8v2",$conn2);
$query2 = "SELECT * FROM tb_did_avalaible";
$result2 = mysql_query($query2,$conn2);
var_dump($result2);
?>
Altough there's no need for 2 connections, you can select both DB's using the same connection.
Sorry i just figure out the problem. If using same connection parameter, must add true in the connect parameter
$conn1 = mysql_connect("localhost","root","passw0rd") or die(mysql_error());
$conn2 = mysql_connect("localhost","root","passw0rd",true) or die(mysql_error());
Don't use mysql connector, use mysqli. It is more secure compared to mysql.
the code would be.
$conn1 = new mysqli("localhost","user","password","db1");
$conn2 = new mysqli("localhost","user","password","db2");
$query1 = "select * from table1";
$query2 = "select * from table2";
echo $query1 . "<br />";
echo $query2 . "<br />";
$rs1 = $conn1->query($query1);
$rs2 = $conn2->query($query1);
Also check if the the query is correct. Most of the times the error is in the query and not the syntax.