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  • Bash loop command until file contains n duplicate entries (lines)

    - by Andrew
    Hello, I'm writing a script and I need to create a loop that will execute same commands until file does contain a specified number of duplicate entries. For example, with each loop I will echo random string to file results. And I want loop to stop when there are 10 lines of of the same string. I thought of something like while [ `some command here (maybe using uniq)` -lt 10 ] do command1 command2 command3 done Do you have any idea how can this problem be solved? Using grep can't be done since I don't know what string I need to look for. Thank you for your suggestions.

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  • How to run a set of SQL queries from a file, in PHP?

    - by Harish Kurup
    I have some set of SQL queries which is in a file(i.e query.sql), and i want to run those queries in files using PHP, the code that i have wrote is not working, //database config's... $file_name="query.sql"; $query==file($file_name); $array_length=count($query); for($i=0;$i<$array_length;$i++) { $data .= $query[$i]; } echo $data; mysql_query($data); it echos the SQL Query from the file but throws an error at mysql_query() function...

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  • Display Wordpress parent Category with their child category

    - by saorabh
    Hi, I want to display only those the parent category which have some child category with their child category without using child_of= I was trying to display but i am only able to get the list of child category not their parent category name. <?php $querystr = "SELECT wp_terms.name, wp_terms.term_id, wp_terms.name FROM wp_terms, wp_term_taxonomy WHERE wp_terms.term_id = wp_term_taxonomy.term_id AND wp_term_taxonomy.parent !=0 "; $cat_child = $wpdb->get_results($querystr, OBJECT); var_dump($cat_child); foreach ($cat_child as $category) { echo $category->name. ' , '; } ?> Help me.. Thanks

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  • Display loading image while post with ajax

    - by DonJoe
    I know there are thousands of examples on the internet, but I want for the script I already have to display a loading gif image while the data is retrievedd. My java knowledge are poor, therefore I'm asking how to change the following: <script type="text/javascript"> $(document).ready(function(){ function getData(p){ var page=p; $.ajax({ url: "loadData.php?id=<? echo $id; ?>", type: "POST", cache: false, data: "&page="+ page, success : function(html){ $(".content").html(html); } }); } getData(1); $(".page").live("click", function(){ var id = $(this).attr("class"); getData(id.substr(8)); }); }); </script> And my div is here: <div class="content" id="data"></div> Thanks. John

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  • How to prevent filename expansion in for loop in bash

    - by cagri
    In a for loop like this, for i in `cat *.input`; do echo "$i" done if one of the input file contains entries like *a, it will, and give the filenames ending in 'a'. Is there a simple way of preventing this filename expansion? Because of use of multiple files, globbing (set -o noglob) is not a good option. I should also be able to filter the output of cat to escape special characters, but for i in `cat *.input | sed 's/*/\\*'` ... still causes *a to expand, while for i in `cat *.input | sed 's/*/\\\\*'` ... gives me \*a (including backslash). [ I guess this is a different question though ]

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  • Simple image displaying php error

    - by Rani dubey
    In a query say i have a image path that have multiple images stored as path=http://localhost/images/xyz.When i run the code: <?php //Retrieves data from MySQL mysql_connect("localhost", "root", "") or die ("Could not save image name Error: " . mysql_error()); mysql_select_db("dawat") or die("Could not select database"); $data = mysql_query("SELECT 'images_path' FROM images_tbl") or die(mysql_error()); //Puts it into an array $file_path = 'http://localhost/images/xyz'; while($row = mysql_fetch_assoc( $data )) {//Outputs the image and other data $src=$file_path.$row["images_path"]; echo "<img src=".$src."> <br>"; } ?> Everything is working fine,but only images are not showing.On place of images it is showing small thumnails.Please suggest what to do.... Sql query i used is: CREATE TABLE images_tbl( images_id INT NOT NULL AUTO_INCREMENT, images_path VARCHAR(200) NOT NULL, submission_date DATE, PRIMARY KEY (images_id) );

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  • PHP Image content type problem

    - by Mirko
    Hi everybody, I have a specific problem, and cant get over it. For my latest project I need a simple PHP script that display an image according to its ID sent through URL. Here's the code: header("Content-type: image/jpeg"); $img = $_GET["img"]; echo file_get_contents("http://www.somesite.hr/images/$img"); The problem is that the image doesn't show although the browser recognizes it (i can see it in the page title), instead I get the image URL printed out. It doesn't work neither on a server with remote access allowed nor with one without. Also, nothing is printed or echoed before the header. I wonder if it is a content type error, or something else. Thanks in advance.

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  • numbers of Parameters in Webservice function

    - by sachin
    hi, I am trying to call webservice from python client using SUDS. as per SUDS support, (https://fedorahosted.org/suds/wiki/Documentation#OVERVIEW) I created a webservice with Config: SOAP Binding 1.1 Document/Literal though Document/literal style takes only one parameter, SUDS Document (https://fedorahosted.org/suds/wiki/Documentation#BASICUSAGE) shows: Suds - version: 0.3.3 build: (beta) R397-20081121 Service (WebServiceTestBeanService) tns="http://test.server.enterprise.rhq.org/" Prefixes (1): ns0 = "http://test.server.enterprise.rhq.org/" Ports (1): (Soap) Methods: addPerson(Person person, ) echo(xs:string arg0, ) getList(xs:string str, xs:int length, ) getPercentBodyFat(xs:string name, xs:int height, xs:int weight) getPersonByName(Name name, ) hello() testExceptions() testListArg(xs:string[] list, ) testVoid() updatePerson(AnotherPerson person, name name, ) Types (23): Person Name Phone AnotherPerson Which has functions with several or no parameters. can we have such methods(Exposed) in a webservice with Document/Literal Style? if so how?

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  • pagination in css/php

    - by fusion
    two questions: --1-- it displays all the number of pages. but i'd like it to display as: << Prev . . 3 4 [5] 6 7 . . Next --2-- when i hover on the current page no, it should change color or increase the font-size, but when i modify the css of a:hover, all the page nos change color or size instead of the one which i'm pointing to. also, when modifying a:active, nothing happens. this is my paging code in php: $self = $_SERVER['PHP_SELF']; $limit = 2; //Number of results per page $numpages=ceil($totalrows/$limit); $query = $query." ORDER BY idQuotes LIMIT " . ($page-1)*$limit . ",$limit"; $result = mysql_query($query, $conn) or die('Error:' .mysql_error()); ?> <div class="caption">Search Results</div> <div class="center_div"> <table> <?php while ($row= mysql_fetch_array($result, MYSQL_ASSOC)) { $cQuote = highlightWords(htmlspecialchars($row['cQuotes']), $search_result); ?> <tr> <td style="text-align:right; font-size:15px;"><?php h($row['cArabic']); ?></td> <td style="font-size:16px;"><?php echo $cQuote; ?></td> <td style="font-size:12px;"><?php h($row['vAuthor']); ?></td> <td style="font-size:12px; font-style:italic; text-align:right;"><?php h($row['vReference']); ?></td> </tr> <?php } ?> </table> </div> <div class="searchmain"> <?php //Create and print the Navigation bar $nav=""; $next = $page+1; $prev = $page-1; if($page > 1) { $nav .= "<span class=\"searchpage\"><a onclick=\"showPage('','$prev'); return false;\" href=\"$self?page=" . $prev . "&q=" .urlencode($search_result) . "\">< Prev</a></span>"; $first = "<span class=\"searchpage\"><a onclick=\"showPage('','1'); return false;\" href=\"$self?page=1&q=" .urlencode($search_result) . "\"> << </a></span>" ; } else { $nav .= "&nbsp;"; $first = "&nbsp;"; } for($i = 1 ; $i <= $numpages ; $i++) { if($i == $page) { $nav .= "<span class=\"searchpage\">$i</span>"; }else{ $nav .= "<span class=\"searchpage\"><a onclick=\"showPage('',$i); return false;\" href=\"$self?page=" . $i . "&q=" .urlencode($search_result) . "\">$i</a></span>"; } } if($page < $numpages) { $nav .= "<span class=\"searchpage\"><a onclick=\"showPage('','$next'); return false;\" href=\"$self?page=" . $next . "&q=" .urlencode($search_result) . "\">Next ></a></span>"; $last = "<span class=\"searchpage\"><a onclick=\"showPage('','$numpages'); return false;\" href=\"$self?page=$numpages&q=" .urlencode($search_result) . "\"> >> </a></span>"; } else { $nav .= "&nbsp;"; $last = "&nbsp;"; } echo $first . $nav . $last; ?> </div> and this is how it displays currently: css: .searchmain { margin:30px; text-align: center; } .searchpage { border: solid 1px #ddd; background: #fff; text-align:left; font-size: 16px; padding:9px 12px; color: #FEBE33; margin-left:2px; } .searchpage a{ text-decoration: none; color: #808080; } .searchpage a:hover { color: #FEBE33; border-color: #036; text-decoration: none; } .searchpage a:visited { color: #808080; }

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  • change a value xml in php but false with a node name id-7

    - by Nataly Nguyen
    I want to change a xml, but fails with this code. I think mistake with name of variable (ID-1) in php. chang.php <?php include 'example.php'; $xml = new SimpleXMLElement($xmlstr); $xml->ID-1 = '8'; $xml->name = 'Big Cliff'; $xml->asXML('test2.xml'); echo $xml->asXML(); ?> example.php <?php $xmlstr = <<<XML <?xml version="1.0" encoding="utf-8"?> <film> <ID-1>29</ID-1> <name>adf</name> </film> XML; ?>

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  • How to show / debug PEAR::DB errors in webpage?

    - by Markus Ossi
    I am connecting to MySQL database on my webpage and have this copy-pasted code for errors: if(DB::isError($db)) die($db->getMessage()); I have the connection code in an outside file called connection.inc that I include at the beginning of my page before the DOCTYPE and html tags. For debugging purposes, how can I print the database errors on my webpage? I thought I could do something like this: echo 'Could not connect to database. The error was:' . $db->getMessage(); but this returns: Fatal error: Call to undefined method DB_mysql::getMessage()

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  • How to begin writing an Application Server over Apache?

    - by Cracker
    For my college project, I want to create a simple application server in C that runs over Apache. Like .php, .asp, .jsp, the extension of my files would be .sas. I have already written a parser which reads the .sas files and generates the output. For example, consider a file index.sas with the below code: <% echo "Hello"; %> Now, if I execute: sas index.sas The result would be: Hello Now I want to use this program as an application server over Apache just as PHP, Tomcat, etc. work over Apache. I have heard of cgi-bin but I think PHP uses a different approach. I want to learn the approach which PHP uses. Please advice.

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  • Pass javascript array to php by using curly braces key name

    - by user7031
    My js code: $(function(){ var arr = new Array('jj', 'kk', 'oo'); $.post('test12.php', {'arr[]': arr}, function(data){ alert(data); }); }); PHP code: <?php echo print_r($_POST['arr']); The thing is,$.post receive a key named 'arr[]',it should be used in PHP as 'arr[]' instead of 'arr',but '$_POST['arr[]']' doesn't work,'arr' works.Which seems that Jquery might do something with curly braces '[]' before sending something to PHP. Secondly,when I remove the single quotas around 'arr[]',PHP can not receive anything by using $_POST['arr'];,I don't know why? Doing this task in a traditional way with no curly braces: $.post('test12.php', {arr: arr}, function(data){ alert(data); }); It works fine. So when sending javascript array to PHP,why bothering using single quote and curly braces like 'arr[]' instead of using a concise way like arr:arr My return result is Array( [0]=>jj [1]=>kk [2]=>oo ) 1 Notice there is a 1 under the array,why?

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  • Magento table rates custom options

    - by Usman Ahmad
    in Tablerate.php I want to change the calculation. So for some Products with custom options like width, height the shipping cost must change. I tried with this method to find out if one product in cart has width or height greater than 60cm (example). But currently I have no Idea how to get custom option values... this code working well. foreach ($request->getAllItems() as $item) { echo 'Name: '.$item->getName(). '<br/> SKU:'.$item->getSku(). '<br/> ProductID: '.$item->getProductId(). '<br/> Price: '.$item->getPrice().'<br/>'; } Thanks

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  • PHP is not returning me a number type

    - by Tristan
    Hello, i tryed to follow that great tutorial (STAR rating with css : http://stackoverflow.com/questions/1987524/turn-a-number-into-star-rating-display-using-jquery-and-css) but i've just a big problem : When i do <span class="stars">1.75</span> or $foo='1.75'; echo '<span class="stars">'.$foo.'</span> the stars is correctly shown, but as soon as i do : while($val = mysql_fetch_array($result)) { $average = ($val['services'] + $val['serviceCli'] + $val['interface'] + $val['qualite'] + $val['rapport'] ) / 5 ; <span class="stars">.$average.</span> } the stars stops working i double checked the data type in mysql : they're all TINYINT(2) and i tryed that : $average = intval($average); but it's still not working, Thank you

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  • why won't my ajax work asynchronously

    - by payling
    I'm having trouble understanding why my code will not work asynchronously. When running asynchronously the get_price.php always receives the same $_GET value even though the alert before outputs a unique $_GET value. var arraySize = "<? echo count($_SESSION['items']); ?>"; //get items count var pos = 0; var pid; var qty; getPriceAjax(); function getPriceAjax() { pid = document.cartItemForm.elements[pos].id; //product id qty = document.cartItemForm.elements[pos].value; //quantity alert('Product: ' + pid + ' Quantity: ' + qty); $.ajax({ url:"includes/ajax_php/get_price.php", type:"GET", data:'pid='+pid+'&qty='+qty, async:true, cache:false, success:function(data){ while(pos < arraySize) { document.getElementById(pid + 'result').innerHTML=data; pos++; getPriceAjax(); } } }) }

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  • node_load in drupal gets incorrect node when you are NOT logged in

    - by Alaa
    Hi All, i have a module and i am using node_load(array('nid' = arg(1))); now the problem is that this function keep getting its data for node_load from DB cache. how can i force this function to not use DB cache or static value? Example my link is http://mydomain.com/node/344983 now: $node=node_load(array('nid'=arg(1)),null,true); echo $node-nid; output: 435632 which is a randomly node id (available in the database) and everytime i ctrl+F5 my browser, i get new nid!! Note: if i am logged in, it gives the result correctly, but this problem happens only when i am browsing the website as an anonymous user i really appreciate any idea!! Thanks

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  • Jquery delay timeout function???

    - by iSimpleDesign
    I am really struglling tring to get this to work what i what is if my php script returns success. echo success I want it to should a message that says congratulations its all setup but stay for aleast 5 seconds but it never seems to work i have tried elay etc but still getting issues please help. here is my code it works but for about a second it then redirects far to quick to read it. if($.trim(data) == 'Congratulations'){ setTimeout(function(){ $('#congrats').fadeIn(1000,function(){ window.location.href='http://example.co.uk/tour/first-time-users'; }); },5500);

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  • Form Security (discussion)

    - by Eray Alakese
    I'm asking for brain storming and sharing experience. Which method you are using for form submiting security ? For example , for block automatically sended POST or GET datas, i'm using this method : // Generating random string <?php $hidden = substr(md5(microtime()) ,"-5"); ?> <form action="post.php" .... // assing this random string to a hidden input <input type="hidden" value="<?php echo $hidden;" name="secCode> // and then put this random string to a session variable $_SESSION["secCode"] = $hidden; **post.php** if ($_POST["secCode"] != $_SESSION["secCode"]) { die("You have to send this form, on our web site"); }

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  • OpenOffice Calc Macro: Run shell command and return output as result of custom function

    - by Mark
    I would like to write a custom OpenOffice function that runs a shell command and puts the result into the cell from which it was invoked. I have a basic macro working, but I can't find a way to capture the command's output. Function MyTest( c1 ) MyTest = Shell("bash -c "" echo hello "" ") End Function The above always returns 0. Looking at the documentation of the Shell command, I don't think it actually returns STDOUT. How would I capture the output so that I can return it in my function? Thanks!

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  • How can I get Syslogging to work on the JVM?

    - by Synesso
    I want to do syslogging from Java. There is a log4j appender, but it doesn't seem to work (for me anyway ... though Google results show many others with this issue still unresolved). I'm trying to debug the appender, so I've written the following script based upon RFC3164 It runs, but no logging appears in the syslog. // scala import java.io._ import java.net._ val ds = new DatagramSocket() val fullMsg = "<11>May 26 14:47:22 Hello World" val packet = new DatagramPacket(fullMsg.getBytes("UTF-8"), fullMsg.length, InetAddress.getLocalHost, 514) ds send packet ds.close I also tried using /bin/nc, but it doesn't work either. echo "<14>May 26 15:23:83 Hello world" > nc -u localhost 514 The Ubuntu command /usr/bin/logger does work, however. logger -p user.info hello world # logs: May 26 15:25:10 dsupport2 jem: hello world What could I be doing wrong?

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  • PHP imap_search not detecting all messages in gmail inbox

    - by Steve
    When I run a very simple imap_search on my GMail inbox, the search returns less messages than it should. Here is the script that anyone with a GMail account can run. $host = '{imap.gmail.com:993/imap/ssl}'; $user = 'foo'; $pass = 'bar'; $imapStream = imap_open($host,$user,$pass) or die(imap_last_error()); $messages = imap_search($imapStream,"ALL"); echo count($messages); imap_close($imapStream); This returns 39 messages. But, I've got 100 messages in my inbox, some in conversations, some forwarded from another account (SquirrelMail, FWIW). Can anyone duplicate these results, and/or tell me what's going on? Other server strings I've tried, all returning the same results: {imap.gmail.com:993/imap/ssl/novalidate-cert} {imap.gmail.com:993/imap/ssl/novalidate-cert}INBOX {imap.gmail.com:993/imap/ssl}INBOX GMail's IMAP feature support: http://mail.google.com/support/bin/answer.py?hl=en&answer=78761

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  • Sending mail from a contact form. Simple but challenging!

    - by ekalaivan
    I've set up a contact form in the Greet Us page in http://swedsb.com When I submit the form, it says mail sent successfully. But I'm not getting the mail, checked the spam folder. I've set a similar form at http://ibsolutions.in. It is working perfectly. Been breaking my head for the past 4 hours. Here's my contact.php <?phpif(isset($_POST['submit'])) { $to = "[email protected]"; $subject = $_POST['posRegard']; $name = $_POST['posName']; $email = $_POST['posEmail']; $message = $_POST['posText']; $body = "$name has sent you a greeting. \n E-Mail: $email\nMessage:\n $message"; mail($to, $subject, $body); header( 'Location: http://swedsb.com/' ); } else {echo "blarg!";}?

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  • JQuery,ajax problem?

    - by user303832
    Hello,I have one table,and when I click on row,it load some content in another table,problem is,when I first time click on row it loads just one time(message 'Some msg' and message 'Some other msg' is showen one time),when I click on other row,it loads twice(messages is shown twice),third time when I click on row it loads three times,ets.Here is my code. $.ajax({ url:'<?php echo $full_path_ajax_php; ?>', data:{'what':'2','mypath':'12345678'}, dataType:'json', type: 'GET', beforeSend:function(){alert('Some msg')}, success: function(){alert('Some other msg')} }); return false; Can someone help me please to understand this.Tnx in advance.

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  • mysql_connect()

    - by Jacksta
    I am trying to connect to mysql and am getting an error. I put my servers ip address in and used port 3306 whihch post should be used? <?php $connection = mysql_connect("serer.ip:port", "user", "pass") or die(mysql_error()); if ($connection) {$msg = "success";} ?> <html> <head> </head> <body> <? echo "$msg"; ?> </body> </html> Here is the error its producing Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'admin'@'server1.myserver.com' (using password: YES) in /home/admin/domains/mydomain.com.au/public_html/db_connect.php on line 3 Access denied for user 'admin'@'server1.myserver.com' (using password: YES)

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