|
PHP - Mysql Connection Details
Perhaps this is a rather lay question, but, is there a way to gather specific connection details about an open MySQL connection in php?
Example: <?php $connect = mysql_connect('localhost', 'username', 'password'); ?> Using the $connect variable, could I run a command that dumps the host, and username to a log file?? Thanks in advance, I'm still searching. E Similar TutorialsI think ive finished the piece of code below, after using escape string for the first time. Ive also put my connection details in a different folder on my hosting account root (worried that this would of been displayed in the event of a parsing eror), is there anything else I can do to make my site secure? Code: [Select] <?php include('func.php'); include($_SERVER['DOCUMENT_ROOT'].'/include/db.php'); ?> <!--$INC_DIR = $_SERVER["DOCUMENT_ROOT"]. "/include/";--> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Chained Select Boxes using PHP, MySQL and jQuery</title> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3/jquery.min.js"></script> <script type="text/javascript"> $(document).ready(function() { $('#wait_1').hide(); $('#drop_1').change(function(){ $('#wait_1').show(); $('#result_1').hide(); $.get("func.php", { func: "drop_1", drop_var: $('#drop_1').val() }, function(response){ $('#result_1').fadeOut(); setTimeout("finishAjax('result_1', '"+escape(response)+"')", 400); }); return false; }); }); function finishAjax(id, response) { $('#wait_1').hide(); $('#'+id).html(unescape(response)); $('#'+id).fadeIn(); } </script> </head> <body> <p> <form action="" method="post"> Name: <input type="text" name="Name" /><br /> Phone: <input type="text" name="Phone" /><br /> Email: <input type="text" name="Email" /><br /> Postcode: <input type="text" name="Postcode" /><br /> Web Address: <input type="text" name="Website" /><br /><br /> <select name="drop_1" id="drop_1"> <option value="" selected="selected" disabled="disabled">Select a Category</option> <?php getTierOne(); ?> </select> <span id="wait_1" style="display: none;"> <img alt="Please Wait" src="ajax-loader.gif"/> </span> <span id="result_1" style="display: none;"></span> <br /> </form> </p> <p> <?php if(isset($_POST['submit'])){ $drop = mysql_real_escape_string($_POST['drop_1']); $tier_two = mysql_real_escape_string($_POST['Subtype']); echo "You selected "; echo $drop." & ".$tier_two; } $Name = mysql_real_escape_string($_POST["Name"]); $Phone = mysql_real_escape_string($_POST["Phone"]); $Email = mysql_real_escape_string($_POST["Email"]); $Postcode = mysql_real_escape_string($_POST["Postcode"]); $Website = mysql_real_escape_string($_POST["Website"]); echo "<br>"; echo $Name; echo "<br>"; echo $Website; $query = ("INSERT INTO business (`id`, `Name`, `Type`, `Subtype`, `Phone`, `Email`, `Postcode`, `Web Address`) VALUES ('NULL', '$Name', '$drop', '$tier_two' , '$Phone', '$Email', '$Postcode', '$Website')"); mysql_query($query) or die ( "<br>Query: $query<br>Error: " .mysql_error()); ?> </body> </html> I have created a button which when pressed should present the user with their details (whoever is logged in), here is the form code: <form id="form1" name="form1" method="post" action="getdetails.php"> <input type="submit" name="Get Details" value="Get Details" /> </label> </p> </form> Here is the getdetails.php file <?php mysql_connect("localhost","root",""); mysql_select_db("test"); $username = $_POST['textfield']; echo '</br>'; $query = mysql_query("SELECT * FROM membersdetails WHERE name=`$username` "); while($result = mysql_fetch_array($query)) { //display echo $result['firstname']; echo $result['surname']; } ?> Its not workin at all I have attacthed the error i am getting Any help please? i want to make a monthly report the user selects month from drop down and i must get the specified dates of that month from the DB I am using ajax to get the dates Hello everyone, I am trying to have a function on my website where the administrator can add a new member to the database. Their details are to be stored in the table memberdetails, I have posted the code below, the error i recieve is "Error: Column count doesn't match value count at row 1" Can anybody help me please? form code: <form action="insert.php" method="post"> Username: <input type="text" name="username" /><br><br> Firstname: <input type="text" name="firstname" /><br><br> Surname : <input type="text" name="surname" /><br><br> Date Birth: <input type="text" name="dob" /><br><br> Total Wins: <input type="text" name="wins" /> Total Loses: <input type="text" name="loses" /><br><br> Email Add: <input type="text" name="email" /><br><br> Country : <input type="text" name="born" /><br><br> Other Info: <input type="text" name="other" /><br><br> <input type="submit" name="Submit" value="Create" align="right"></td> </form> insert.php <?php mysql_connect ("localhost","root","") or die("Cannot connect to Database"); mysql_select_db ("test"); $sql="INSERT INTO memberdetails (username, firstname, surname, dob, totalwins, totalloses, email, country, info) VALUES ('$_POST[username]''$_POST[firstname]','$_POST[surname]','$_POST[wins]''$_POST[loses]''$_POST[email]''$_POST[born]''$_POST[other]''$_POST[dob]')"; if (!mysql_query($sql)) { die('Error: ' . mysql_error()); } echo "1 record added"; ?> hi, i'm new in php/mysql. i'm stored student marks values in following format in mysql db table. id student_code Tamil English Maths Science Social 1 1 100 75 78 88 95 2 2 85 90 88 80 100 But i want to search and display the specific student marks in following format. id:1 student_code:1 Tamil:100 English:75 Maths:78 Science:88 Social:92 Total:? Avg:? please give correct code for this format. Obviously when connecting to php Im not going to show all of my login details; mysql_connect("details","details","password") or die(mysql_error()); mysql_select_db("details") or die(mysql_error()); whats the best way to hide them? Ive seen some people using an include file with their login details on but say for eg. <?php include('con.php'); ?> Whats to stop somone looking at www.myweb/con.php and obtaining my details there instead? hi i had database with field of name,title,post,content i want to fetch the post and content for a specific user from giving name of that user by form help me to get that ps just give me idea to how to do that/ Code: [Select] <form id="form1" name="form1" method="post" action="view.php"> <label>Name <input type="text" name="textfield" /> </label> <p> <label> <input type="submit" name="Submit" value="Submit" /> </label> </p> </form> Hi, I have successfully implemented a master details page with the results aligned in columns linking to a details page. I wish to maintain the recordID passed from the master details page and make the dynamic text, which reads Shade A tree that is capable of..... in the attached screen shot a link to another details page referencing the same recordID. The detailspage2.php would look the same as the screenshot except the Shade text and description below will be highlighted, which I can do, there will be a new image and a new image description. All other dynmaic elements on the page will remain the same. I tried to simply save as my detailspage.php to detailspage2.php and create a link to detailspage2.php. It linked to detailspage2.php but none of the record info showed up in their respective table cells. I have all the names desc's, images, etc setup in a table in my database. Please let me know what code and other info you need to help me out with this procedure. Thanks. Hi All, I've searched long and hard accross the web for an answer to this and finnally given in and requesting help. Here's what i have, i have a database setup and working fine. What i would like to do is for an administrator to be able to update my users details. It may sound odd, why don't you let your users update their own details? Well the administrators are dispatchers if you like, and my users are the 'dispatchees', for want of a better word. So i would like my administrators to be able to dispatch my users with routes and my users be able to see the routes that have been dispatched to them. I've setup a login area and a page that pulls there routes off the database, depending on their login details, i.e. jack will see his routes and jill will see her's independantly. This works by me editing the appropriate columns/rows of my database using phpmyadmin. What i'd like now is for administrators (who are directed to a seperate page, with more controls) to be able to do the same as me (updating the database) but by using a php form/script. I'd like to be able to select the routes from a second table on the same database if possible, to try and keep everything tidy. So my dispatcher would select Route001 from a drop down list, this would fill in the text fields next to the route field with From To, so my dispatcher would know what route001 actually is from/ too, choose a username (now being driven from my other table) and hit dispatch. My user would login to their area, hit view dispatched routes and it would display Route 001 with the correct information. The login area was a downloaded script i modified to suit and is called Login-Redirect_v1.31_FULL Many thanks in advance, hope you can sort of understand what i want Josh PHP/MySQL ability:Novice hi everyone I have a question... does creating a connection to mysql takes time depeding on the database size? Lets say that in my entire project I will be connection to 4 databases so I created a config.php which looks like this: Code: [Select] $hostname = "localhost"; $username = "root"; $pword = ""; $con1 = @mysql_connect($hostname,$username,$pword,true); @mysql_selectdb("databasename1",$con1); $con2 = @mysql_connect($hostname,$username,$pword,true); @mysql_selectdb("databasename2",$con2); $con3 = @mysql_connect($hostname,$username,$pword,true); @mysql_selectdb("databasename3",$con3); $con4 = @mysql_connect($hostname,$username,$pword,true); @mysql_selectdb("databasename4",$con4); so, this file is included in every pages, I put this on the top. Basically every page request it will open 4 connection and the script will only use 2 connection and another page will only 1 and so on..... this style is very convenient as for I am not creating a connection in every page.. But my concern is will it effect the performance of my system? Tnx in advance..... I am slightly puzzled when I attempt to connect to MySQL using variables, if I use this:- Code: [Select] // Connect to MySQL and select dbase mysql_connect("$host", "$user", "$pwd") or die("Computer say NO!"); mysql_select_db("test_site") or die("I've just had a senior moment, and cannot connect to the database..."); I get this error:- Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'root'@'localhost' (using password: YES) in C:\wamp\www\php\test site\regproc.php on line 24 Computer say NO! Whereas if I use this:- Code: [Select] // Connect to MySQL and select dbase mysql_connect("localhost", "root", "my password") or die("Computer say NO!"); mysql_select_db("test_site") or die("I've just had a senior moment, and cannot connect to the database..."); it works perfectly. Most of the tutorials I have read seem to indicate that one can use variables in this case, so I do not understand why it doesn't work... Hello guys. Trying to connect php with mysql database and then display results on the screen. This is my code: Code: [Select] <?php $dbhost = "localhost"; $dbuser = "username1"; $dbpass = "password1"; $db = "username1_myDB"; $connection = mysql_connect($dbhost, $dbuser, $dbpass) or die ("Could not connect"); mysql_select_db($connection, $db); $show = "SELECT Name, Description FROM people"; $result = mysql_query($show); while($show = mysql_fetch_array($result)){ $field01 = $show[Name]; $field02 = $show[Description]; echo "id: $field01<br>"; echo "description: $field02<p>"; } ?> However im getting this: Warning: mysql_select_db() expects parameter 1 to be string, resource given in /home/pain33/public_html/index.php on line 20 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/pain33/public_html/index.php on line 26 Any ideas how to fix this? Thank you. Hey there, total n00b here... Just installed and learned the basic functions of PHP and MySQL (Only have very basic programming background.) Also managed to get an Apache server running. I'm attaching a screenshot that should explain my problem entirely but in short... I can connect to and send SQL queries to my MySQL server using my terminal. My browser reads php statements over the apache server fine and it 'seems' to connect to my MySQL server fine, but it won't select a database and doesn't seem to accept other SQL queries. If I use a non-existant server name I get an error code but when I change to a non-existant user name I don't get an error code. So I'm not sure if it actually does connect or not or why the other queries doesn't work. As I said I'm VERY new to this, so I might be making some other mistake as well. Please see attached document/screenshot. Thank you Ardinent I am trying to get data out off a mysql database, by using an php file. But i am not getting any output?
Examples.html
<html ng-app="countryApp">
<head>
<meta charset="utf-8">
<title>Angular.js Example</title>
<script src="http://cdnjs.cloudflare.com/ajax/libs/angular.js/1.2.1/angular.min.js"></script>
<script>
var countryApp = angular.module('countryApp', []);
countryApp.controller('CountryCtrl', function ($scope, $http){
$http.get('category.php').success(function(data) {
$scope.countries = data;
});
});
</script>
</head>
<body ng-controller="CountryCtrl">
<table>
<tr ng-repeat="country in countries">
<td>{{country}}</td>
</tr>
</table>
</body>
</html>
category.php
<?php
$servername = "localhost:3306";
$username = "root";
$password = "root";
$dbname = "myDB";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if(!$conn){
die("Connection failed: " .mysqli_connecet_error());
}
$showData = "SELECT id FROM myDB";
$data = array();
$result = mysqli_query($conn, $showData);
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_assoc($result)){
$data[] = $row;
}
} else {
echo "0 results";
};
print json_encode($data);
mysqli_close($conn);
echo($outp);
?>
Hi Friends, i am new one to this.i have a big problem i think i will get the correct solution... when i am connecting to the MySql Database i am getting this. Warning: mysql_connect() [function.mysql-connect]: Lost connection to MySQL server at 'reading initial communication packet', system error: 110 in /home/txtimg0/public_html/cherple/Scheduled_Campaigns.php on line 24 unable to select database define("prod_dbuser","xxxx"); //gtmdev1 define("prod_dbpass","xxxxx"); //gtmdev1 define("prod_database_cm2","cm2"); define("prod_database_glue","glue"); define("prod_hostname","xxxxxxxx"); i am defined these constants in gtm _constants.php i am includeing this file when ever i need to connect with database. this working fine in past.now we installed the new Database from then i am getting this error. can any one Please me Help to Resolve this Issue.? and why php is not connecting to the database??? Thanks, Ramky I have a class and a lot of the function in the class require a connection to a MySQL database. At the moment I am using the "__construct()" to connect to the database each time a new instance of the classes is created. Is this the best way to do it or should I have a separate class for connection to the database? Just wanted to get an idea of what other people do. Thanks for any help. Hi,
If I place the following into a file called db.php and save it into folder called includes using xampp:
$dbhost = 'localhost'; I'm suddenly having trouble using my connection to my MySQL database... (yes it was working but now...) I have the Connection created in an include file and stored in variable $DB, in the main file that includes the file containing the$DB there are other includes for classes. These classes are SUPPOSED to use $DB to connect to and SELECT/UPDATE/INSERT, but for a reason I cant figure out they suddenly stopped seeing $DB. it keeps saying its an undefined variable. If you need to see code I can post... The below scenario works fine but why shouldnt you do this? Why can I not just run the mysql_connect and mysql_select_db at the top of each page and then run my queries under all of this. What I have read is that the mysql_connection will die after the script finishes (at the end of the page) any way so why do people create mysql_connection objects as php is stateless so it doesnt save this any way. Code: [Select] <?php mysql_connect('localhost', 'user', 'pass'); mysql_select_db('test'); ?> <html> <body> <?php $query = mysql_query("Select * from data"); while($row = mysql_fetch_array($query)) { echo $row['name']."<br />"; } ?> -------------- OTHER HTML HERE --------------- <?php $query = mysql_query("Select * from addresses"); while($row = mysql_fetch_array($query)) { echo $row['postcode']."<br />"; } ?> </body> </html> |
|||||||