Search Results

Search found 18661 results on 747 pages for 'linq to mysql'.

Page 518/747 | < Previous Page | 514 515 516 517 518 519 520 521 522 523 524 525  | Next Page >

  • Selecting Date Range on a PHP form and displaying results from MySQL database

    - by Sarah HSL
    This may be something simple but I cant understand why this wouldn't work.. I have a php form where you can select a date range from drop downs. I've given the field names day, month year, and day1, month1, year1. When clicking submit it takes you to a second php form. Here is the code for second form: <?php $username="***"; $password="***"; $database="****"; mysql_connect('localhost',$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $day = $_GET['day']; $month = $_GET['month']; $year = $_GET['year']; $day1 = $_GET['day1']; $month1 = $_GET['month1']; $year1 = $_GET['year1']; $date1 = "$year-$month-$day"; $date2 = "$year1-$month1-$day1"; $query = "SELECT * FROM main_stock WHERE curr_timestamp BETWEEN '$date1' AND '$date2'"; $result=mysql_query($query); $num=mysql_num_rows($result); ?> <table border="1" cellspacing="2" cellpadding="2"> <tr> <td><b><font face="Arial, Helvetica, sans-serif">Product Description</font></b></td> <td><b><font face="Arial, Helvetica, sans-serif">Category</font></b></td> <td><b><font face="Arial, Helvetica, sans-serif">Master Category</font></b></td> <td><b><font face="Arial, Helvetica, sans-serif">Barcode</font></b></td> <td><b><font face="Arial, Helvetica, sans-serif">Status</font></b></td> <td><b><font face="Arial, Helvetica, sans-serif">TimeStamp</font></b></td> <td><b><font face="Arial, Helvetica, sans-serif">New Own</font></b></td> <td><b><font face="Arial, Helvetica, sans-serif">Serial No.</font></b></td> </tr> <?php $i=0; while ($i < $num) { $f1=mysql_result($result,$i,"product_desc"); $f2=mysql_result($result,$i,"category"); $f3=mysql_result($result,$i,"mastercategory"); $f4=mysql_result($result,$i,"barcode"); $f5=mysql_result($result,$i,"status"); $f6=mysql_result($result,$i,"curr_timestamp"); $f7=mysql_result($result,$i,"newown"); $f8=mysql_result($result,$i,"serial"); ?> <tr> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f1; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f2; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f3; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f4; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f5; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f6; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f7; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f8; ?></font></td> </tr> <?php $i++; } $num_rows = mysql_num_rows($result); echo "$num_rows Rows\n"; mysql_close(); ?> Is there any reason this wouldn't work? I'm not sure where I am going wrong. It displays results when there is another option as well as the date such as 'status' but when this is taken out and I just want to display all the results between the date range it doesn't work.. This works: <?php $username="+++"; $password="+++"; $database="+++"; mysql_connect('localhost',$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $day = $_GET['day']; $month = $_GET['month']; $year = $_GET['year']; $day1 = $_GET['day1']; $month1 = $_GET['month1']; $year1 = $_GET['year1']; $status = $_GET['status']; $date1 = "$year-$month-$day"; $date2 = "$year1-$month1-$day1"; $query = "SELECT * FROM main_stock WHERE status = '$status' AND curr_timestamp BETWEEN '$date1' AND '$date2'"; $result=mysql_query($query); $num=mysql_num_rows($result); ?> <table border="1" cellspacing="2" cellpadding="2"> <tr> <td><b><font face="Arial, Helvetica, sans-serif">Product Description</font></b></td> <td><b><font face="Arial, Helvetica, sans-serif">Category</font></b></td> <td><b><font face="Arial, Helvetica, sans-serif">Master Category</font></b></td> <td><b><font face="Arial, Helvetica, sans-serif">Barcode</font></b></td> <td><b><font face="Arial, Helvetica, sans-serif">Status</font></b></td> <td><b><font face="Arial, Helvetica, sans-serif">TimeStamp</font></b></td> <td><b><font face="Arial, Helvetica, sans-serif">New Own</font></b></td> <td><b><font face="Arial, Helvetica, sans-serif">Serial No.</font></b></td> </tr> <?php $i=0; while ($i < $num) { $f1=mysql_result($result,$i,"product_desc"); $f2=mysql_result($result,$i,"category"); $f3=mysql_result($result,$i,"mastercategory"); $f4=mysql_result($result,$i,"barcode"); $f5=mysql_result($result,$i,"status"); $f6=mysql_result($result,$i,"curr_timestamp"); $f7=mysql_result($result,$i,"newown"); $f8=mysql_result($result,$i,"serial"); ?> <tr> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f1; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f2; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f3; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f4; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f5; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f6; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f7; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f8; ?></font></td> </tr> <?php $i++; } $num_rows = mysql_num_rows($result); echo "$num_rows Rows\n"; mysql_close(); ?> But when the 'status' field is taken out (and obviously the serial drop down in the first form) it stops working...

    Read the article

  • How to convert my mysql data into this?

    - by sky
    <script type="text/javascript"> var cityList = new Array(....etc), signImgList = new Array('http:\/\/pic.sitename.com:80\/file\/11\/34\/01\/22\/default\/SIGN11340122_48x48.jpg?t=1257391453468'....etc), titleList = new Array(....etc), userSex = new Array(....etc), userAge = new Array('19','26'....etc), userMid = new Array('lwowl','kylin0621'....etc); </script>

    Read the article

  • Select count() max() Date HELP!!! mysql oracle

    - by DAVID
    Hi guys i have a table with shifts history along with emp ids im using this code to retrieve a list of employees and their total shifts by specifying the range to count from: SELECT ope_id, count(ope_id) FROM operator_shift WHERE ope_shift_date >=to_date( '01-MAR-10','dd-mon-yy') and ope_shift_date <= to_date('31-MAR-10','dd-mon-yy') GROUP BY OPE_ID which gives OPE_ID COUNT(OPE_ID) 1 14 2 7 3 6 4 6 5 2 6 5 7 2 8 1 9 2 10 4 10 rows selected. NOW how do i choose the employee with the highest no of shifts under the specified range date, please this is really important

    Read the article

  • Calendar using Javascript/ PHP/ mySQL

    - by Gushiken
    for a current webapp i need a "outlook-like" calendar... Here are some requirements for the calendar: week-view for the appointments different appointment types direct display of the length and time of the date (like in googleCalendar) multiple appointments for the same time only using javascript, php and any DB We need the calendar for the Zend Framework, so if the Calendar doesn't already support the ZF, the source needs to be editable! do you know any calendar which fits my needs? or do you have any tipps for developing one by myself?

    Read the article

  • rails: best way to store comments in mysql

    - by ciss
    Hello. Okay i have two models: posts and comments. as you can think comments has column :post_id. My models Comments belongs_to :post Post has_many :comments So, this is pretty simple association but i have some problems with ordering comments. at first time, when i create my comments migration file i just add column :position. This column indicate comment position in the post. But now i think what where is more good way to do this. so i can't make my choise: 1) uses t.column :datatime :created_at, :default = Time.now() 2) or use timestamps? this is undiscovered for me, please tell me about your exp.

    Read the article

  • how to print data in a table from mysql

    - by robertdd
    hello, i want to extract the last eight entries from my database and print them into a 2 columns table!like this: |1|2| |3|4| |5|6| |7|8| is that possible? this is my code: $db = new Database(DB_SERVER, DB_USER, DB_PASS, DB_DATABASE); $db->connect(); $sql = "SELECT ID, movieno FROM movies ORDER BY ID DESC LIMIT 8 "; $rows = $db->query($sql); print '<table width="307" border="0" cellspacing="5" cellpadding="4">'; while ($record = $db->fetch_array($rows)) { $vidaidi = $record['movieno']; print <<<END <tr> <td> <a href="http://www.youtube.com/watch?v=$vidaidi" target="_blank"> <img src="http://img.youtube.com/vi/$vidaidi/1.jpg" width="123" height="80"></a> </td> </tr> END; } print '</table>';

    Read the article

  • Image edit and mysql

    - by Felicita
    I have a simple table for reference page: id name description image In reference.php, A form upload image to a folder and save image's name in image section. In reference.php?action=edit page I want to edit the image. What is correct way to edit? Uploading another image and update the table? Thanks

    Read the article

  • insert mysql record with table format in a string then echo in a html

    - by user1292042
    Ok, I have this code. $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("juliver", $con); $result = mysql_query("SELECT * FROM items"); while($row = mysql_fetch_array($result)) { $hi1='<img src="'.$row['name'].'" />'; $hi2= $row['title']; $hi3= $row['description']; $hi4= $row['link']; } Now, Im in a process of making those records above to display in a table view and that 4 row should be on one td and so the rest.

    Read the article

  • SQLite MYSQL character encoding

    - by Lee Armstrong
    I have a strange situation where the following code works however XCode warns it is deprecated... NSString *col1 = [NSString stringWithCString:(char *)sqlite3_column_text(compiledStatement, 0)]; However as that is the deprecated method if I set an encoding the string comes out wrong! I have tried all the encodings but none work! NSString *col1 = [NSString stringWithCString:(char *)sqlite3_column_text(compiledStatement, 0) encoding:NSASCIIStringEncoding];

    Read the article

  • News Portals and mysql queries

    - by jasmine
    Consider a news portal like cnn. There is 8 categories. Is there 8 query in every page loading? For eg SELECT title FROM category where cID = 1 SELECT title FROM category where cID = 2 ................... And for sites with high traffic, would be memcache a good solution? Thanks in advance

    Read the article

  • Change HTML DropDown Default Value with a MySQL value

    - by fzr11017
    I'm working on a profile page, where a registered user can update their information. Because the user has already submitted their information, I would like their information from the database to populate my HTML form. Within PHP, I'm creating the HTML form with the values filled in. However, I've tried creating an IF statement to determine whether an option is selected as the default value. Right now, my website is giving me a default value of the last option, Undeclared. Therefore, I'm not sure if all IF statements are evaluation as true, or if it is simply skipping to selected=selected. Here is my HTML, which is currently embedded with PHP(): <select name="Major"> <option if($row[Major] == Accounting){ selected="selected"}>Accounting</option> <option if($row[Major] == Business Honors Program){ selected="selected"}>Business Honors Program</option> <option if($row[Major] == Engineering Route to Business){ selected="selected"}>Engineering Route to Business</option> <option if($row[Major] == Finance){ selected="selected"}>Finance</option> <option if($row[Major] == International Business){ selected="selected"}>International Business</option> <option if($row[Major] == Management){ selected="selected"}>Management</option> <option if($row[Major] == Management Information Systems){ selected="selected"}>Management Information Systems</option> <option if($row[Major] == Marketing){ selected="selected"}>Marketing</option> <option if($row[Major] == MPA){ selected="selected"}>MPA</option> <option if($row[Major] == Supply Chain Management){ selected="selected"}>Supply Chain Management</option> <option if($row[Major] == Undeclared){ selected="selected"}>Undeclared</option> </select>

    Read the article

  • [PHP] converting mysql data into array?

    - by Mahmoud
    i have an ajax script that check if the user name is available or not, but it keeps taking the newest user name and the rest are out $result = mysql_query("Select username from customer"); while ($row = mysql_fetch_array($result)){ $existing_users=array(''.$row['username'].','); } i know i am doing something worng

    Read the article

  • pagination using php with mysql data

    - by pradeep
    i am trying to create a pagination in which there are 5 items at a time depending on the number of items in DB. i wrote this code . but i dono how to go further .its buggy..any better pagination or alteration for this <?php $myresult .= "<div class='pagination' >"; if ($pagenum == 1) { } else { $pagenum = 1; $myresult .= "<a href='javascript:newPage(\"".$pagenum."\")'>&nbsp;first&nbsp;</a>"; $myresult .= " "; $previous = $pagenum-1; $myresult .= "<a href='javascript:newPage(\"".$previous."\")'> &nbsp;Prev&nbsp;</a>"; if ($pagenum == $last) { $previous3 = $pagenum-4; $myresult .= "<a href='javascript:newPage(\"".$previous3."\")'> &nbsp; $previous3 &nbsp;</a>"; $previous2 = $pagenum-3; $myresult .= "<a href='javascript:newPage(\"".$previous2."\")'> &nbsp; $previous2 &nbsp;</a>"; } if ($pagenum > 2) { $previous1 = $pagenum-2; $myresult .= "<a href='javascript:newPage(\"".$previous1."\")'> &nbsp; $previous1 &nbsp;</a>"; } if ($pagenum > 1) { $previous2 = $pagenum-1; $myresult .= "<a href='javascript:newPage(\"".$previous2."\")'> &nbsp; $previous2 &nbsp;</a>"; $myresult .= " "; } } $myresult .= "<span class=\"disabled\"> $pagenum </span>"; if ($pagenum == $last) { } else { if($pagenum < $last - 1) { $next = $pagenum+1; $myresult .= "<a href='javascript:newPage(\"".$next."\")'>&nbsp; $next &nbsp;</a>"; } if($pagenum < $last - 2) { $next1 = $pagenum+2; $myresult .= "<a href='javascript:newPage(\"".$next1."\")'> &nbsp; $next1 &nbsp;</a>"; } if($pagenum == 1 ) { $next2 = $pagenum+3; $myresult .= "<a href='javascript:newPage(\"".$next2."\")'> &nbsp; $next2 &nbsp;</a>"; $next3 = $pagenum+4; $myresult .= "<a href='javascript:newPage(\"".$next3."\")'> &nbsp; $next3 &nbsp;</a>"; } if($pagenum == 2 ) { $next2 = $pagenum+3; $myresult .= "<a href='javascript:newPage(\"".$next2."\")'> &nbsp; $next2 &nbsp;</a>"; } $next = $pagenum+1; $myresult .= "<a href='javascript:newPage(\"".$next."\")'>&nbsp;Next&nbsp;</a>"; $myresult .= "<a href='javascript:newPage(\"".$last."\")'>&nbsp;Last</a>"; } $myresult .= "</div>"; $myresult .= "</br>"; ?>

    Read the article

  • MySql php: check if Row exists

    - by Jeff
    This is probably an easy thing to do but I'm an amateur and things just aren't working for me. I just want to check and see if a row exists where the $lectureName shows. If a row does exist with the $lectureName somewhere in it, I want the function to return "assigned" if not then it should return "available". Here's what I have. I'm fairly sure its a mess. Please help. function checkLectureStatus($lectureName) { $con = connectvar(); mysql_select_db("mydatabase", $con); $result = mysql_query("SELECT * FROM preditors_assigned WHERE lecture_name='$lectureName'"); while($row = mysql_fetch_array($result)); { if (!$row[$lectureName] == $lectureName) { mysql_close($con); return "Available"; } else { mysql_close($con); return "Assigned"; } } When I do this everything return available, even when it should return assigned.

    Read the article

  • fetching only new rows from mysql with jquery ajax

    - by testkhan
    i have a table named news with 3 fields i.e (id, news, time) and i have a setInterval after every 3mints to fetch news from google or any news site .... now i want to fetch only new rows inserted after every 5 minutes...with jquery $.ajax()...how can i do that... do i reload the whole table or there is a way to fetch only the new ones...

    Read the article

  • How to pass values from array into mysql with php

    - by moustafa
    my original code is this <tr> <th> <label for="user_level"> User Level: * <?php echo isset($valid_user_level) ? $valid_user_level : NULL; ?> </label> </th> </tr> <td> <select name="user_level" id="user_level" class="sel"> <option value="">Select one…</option> <option value="1">User</option> <option value="5">Admin</option> </select> </td> this give me the option to select one of choice from the drop down menu i.e. user and when user is selected and the submit button is pressed this will insert the value 1 into the database which will when the user logs in tell the system that they are are normal user. I want to change the code to the following <tr> <td> <select name="user_level" id="user_level" class="sel"> <option value="">Select one…</option> <?php if(!empty($level)) { foreach($level as $value) { echo "<option value='{$value}'"; echo getSticky(2,'user_level',$value); echo ">{$value}</option>"; } } ?> </select> </td> </tr> With this being my array query $level = array('User','Admin'); How can I pass the values of 1 for user level and 5 for admin in this code so when the user is selected it inouts 1 into the database?

    Read the article

  • zend like mysql problem

    - by pradeep
    hi, I am trying to use like in zend switch($filter2) { case 'name': switch($filter1) { case 'start_with': $search = "\"pd_name like ?\", '$patient_search_name%'"; break; case 'contains': $search = "'pd_name like ?', '%$patient_search_name%'"; break; case 'exact_match': $search = "'pd_name = ?', $patient_search_name"; break; } break; case 'phone': switch($filter1) { case 'start_with': $search = "'pd_phone like ?', '$patient_search_name%'"; break; case 'contains': $search = "'pd_phone like ?', '%$patient_search_name%'"; break; case 'exact_match': $search = "'pd_phone = ?', $patient_search_name"; break; } break; } $select = $this->getDbTable()->select() ->from("patient_data", array('*')) ->where("$search"); but when i see the query log its like SELECT `patient_data`.* FROM `patient_data` WHERE ("pd_name like ?", 'bhas%') where as the ? should have been replaced by the value ....how to solve this??

    Read the article

  • Grabbing the right record in a drop down from mysql

    - by 86Stang
    I'm using a form where the user can edit an entry. Everything is populating and all is well with the exception that I can't get the drop down to show the project that they've already assigned the image to. $project_qry = "SELECT * from projects ORDER BY title ASC"; $project_res = mysql_query($project_qry); $project_drop = "<select name=\"project_id\">\n"; while ($row = mysql_fetch_array($project_res)) { if ($project_id == $row[title]) { $project_drop .= "<option value=\"$row[id]\" selected>$row[title]</option>\n"; } else { $project_drop .= "<option value=\"$row[id]\">$row[title]</option>\n"; } } $project_drop .= "</select>\n"; I'm sure it's something devilishly simple but I'm stumped.

    Read the article

  • ReSharper 5.0s LINQ Refactoring Continues to be Amazing!

      In this post, well take a straight forward procedure based set of code and convert it to LINQ using a ReSharper from JetBrains suggestion.   The initial code is as follows: int... This site is a resource for asp.net web programming. It has examples by Peter Kellner of techniques for high performance programming...Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

    Read the article

  • How to get distinct values from the List&lt;T&gt; with LINQ

    - by Vincent Maverick Durano
    Recently I was working with data from a generic List<T> and one of my objectives is to get the distinct values that is found in the List. Consider that we have this simple class that holds the following properties: public class Product { public string Make { get; set; } public string Model { get; set; } }   Now in the page code behind we will create a list of product by doing the following: private List<Product> GetProducts() { List<Product> products = new List<Product>(); Product p = new Product(); p.Make = "Samsung"; p.Model = "Galaxy S 1"; products.Add(p); p = new Product(); p.Make = "Samsung"; p.Model = "Galaxy S 2"; products.Add(p); p = new Product(); p.Make = "Samsung"; p.Model = "Galaxy Note"; products.Add(p); p = new Product(); p.Make = "Apple"; p.Model = "iPhone 4"; products.Add(p); p = new Product(); p.Make = "Apple"; p.Model = "iPhone 4s"; products.Add(p); p = new Product(); p.Make = "HTC"; p.Model = "Sensation"; products.Add(p); p = new Product(); p.Make = "HTC"; p.Model = "Desire"; products.Add(p); p = new Product(); p.Make = "Nokia"; p.Model = "Some Model"; products.Add(p); p = new Product(); p.Make = "Nokia"; p.Model = "Some Model"; products.Add(p); p = new Product(); p.Make = "Sony Ericsson"; p.Model = "800i"; products.Add(p); p = new Product(); p.Make = "Sony Ericsson"; p.Model = "800i"; products.Add(p); return products; }   And then let’s bind the products to the GridView. protected void Page_Load(object sender, EventArgs e) { if (!IsPostBack) { Gridview1.DataSource = GetProducts(); Gridview1.DataBind(); } }   Running the code will display something like this in the page: Now what I want is to get the distinct row values from the list. So what I did is to use the LINQ Distinct operator and unfortunately it doesn't work. In order for it work is you must use the overload method of the Distinct operator for you to get the desired results. So I’ve added this IEqualityComparer<T> class to compare values: class ProductComparer : IEqualityComparer<Product> { public bool Equals(Product x, Product y) { if (Object.ReferenceEquals(x, y)) return true; if (Object.ReferenceEquals(x, null) || Object.ReferenceEquals(y, null)) return false; return x.Make == y.Make && x.Model == y.Model; } public int GetHashCode(Product product) { if (Object.ReferenceEquals(product, null)) return 0; int hashProductName = product.Make == null ? 0 : product.Make.GetHashCode(); int hashProductCode = product.Model.GetHashCode(); return hashProductName ^ hashProductCode; } }   After that you can then bind the GridView like this: protected void Page_Load(object sender, EventArgs e) { if (!IsPostBack) { Gridview1.DataSource = GetProducts().Distinct(new ProductComparer()); Gridview1.DataBind(); } }   Running the page will give you the desired output below: As you notice, it now eliminates the duplicate rows in the GridView. Now what if we only want to get the distinct values for a certain field. For example I want to get the distinct “Make” values such as Samsung, Apple, HTC, Nokia and Sony Ericsson and populate them to a DropDownList control for filtering purposes. I was hoping the the Distinct operator has an overload that can compare values based on the property value like (GetProducts().Distinct(o => o.PropertyToCompare). But unfortunately it doesn’t provide that overload so what I did as a workaround is to use the GroupBy,Select and First LINQ query operators to achieve what I want. Here’s the code to get the distinct values of a certain field. protected void Page_Load(object sender, EventArgs e) { if (!IsPostBack) { DropDownList1.DataSource = GetProducts().GroupBy(o => o.Make).Select(o => o.First()); DropDownList1.DataTextField = "Make"; DropDownList1.DataValueField = "Model"; DropDownList1.DataBind(); } } Running the code will display the following output below:   That’s it! I hope someone find this post useful!

    Read the article

  • Error while installing Xtrabackup

    - by Olin
    I'd like to install XtraBackup (rpm -i percona-xtrabackup-2.1.9-744.rhel6.x86_64.rpm). During the rpm install it told me that it misses a dependency. error: Failed dependencies: perl(DBD::mysql) is needed by percona-xtrabackup-2.1.9-744.rhel6.x86_64 perl(Time::HiRes) is needed by percona-xtrabackup-2.1.9-744.rhel6.x86_64 Then I run yum install perl-Time-HiRes, and yum install perl-DBD-MySQL. For install perl-TImes-Hires has successful but not for perl-DBD-MySQL. Error: file /usr/share/mysql/ukrainian/errmsg.sys from install of mysql-libs-5.1.73-3.el6_5.x86_64 conflicts with file from package MySQL-server-5.6.10-1.el6.x86_64. I also had try to install : yum install cpan cpan DBI cpan DBD::mysql But still get the same error. So I hope someone can explain me what the right fix is, to get XtraBackup running on MySQL.

    Read the article

  • Bugzilla email issue

    - by xian
    My bugzilla system keep hit the following error: There was an error sending mail from '[email protected]' to '[email protected]':Can't send data I think that is some problem with my setting and configuration. First is the urlbase I have tried setting it to bugzilla.example.com, and http://127.0.0.1:81/, and http://10.0.0.236/ (My laptop IP address, I use this laptop to set up bugzilla) but the error still persists. Actually what should I put in the urlbase field? Parameter = Email Under mail_delivery_method, i choose SMTP. Under mailfrom, I put bugzilla-daemon. smtpserver, I tried leaving it blank, or setting it to 220.181.12.12 before, but could not solve my problem For my sql, the following is the data and command I used: C:\mysql\bin>mysql --user=root -p mysql Enter password: 1234 (When I install mysql into my laptop, it ask me to key an username and password, i have key in username as 'cvuser' and password as '1234', but here never ask me to key in any username) Welcome to the MySQL monitor. Commands end with ; or \g. Your MySQL connection id is 1 Server version: 5.5.15 MySQL Community Server (GPL) mysql> GRANT ALL PRIVILEGES ON bugs.* TO 'bugs'@'localhost' IDENTIFIED BY '123456'; Query OK, 0 rows affected (0.03 sec) In C:\Bugzilla\localconfig, I put the following info: # # How to access the SQL database: # $db_host = "localhost"; # where is the database? $db_port = 3306; # which port to use $db_name = "bugs"; # name of the MySQL database $db_user = "bugs"; # user to attach to the MySQL database # # Enter your database password here. It's normally advisable to specify # a password for your bugzilla database user. # If you use apostrophe (') or a backslash (\) in your password, you'll # need to escape it by preceding it with a \ character. (\') or (\\) # $db_pass = '123456'; Can someone tell me where my mistake is? I have googled for this issue for few days but still cannot find the solution.

    Read the article

< Previous Page | 514 515 516 517 518 519 520 521 522 523 524 525  | Next Page >