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PHP - Second Problem With Date Modification.
I am trying to get this script to display the date like Dec 17, 2010 and have the search function work too. If I try to change the date the search malfunctions and if I take the date function away the date is 2010 12 17 I thought we had this nailed down the other day but in my excitement I neglected to test the city search. Here is the script. <?php $find = trim($_GET['find']); $field = $_GET['field']; if($find && $field) { // we have search form submitted // check for values to prevent sql injection $valid_fields = array("venue_state", "venue_city", "start_date"); if(!in_array($field, $valid_fields)) die("Error: Invalid field!"); //connect mysql_connect("localhost", "arts_cshow", "TrPh123Yuo") or die(mysql_error()); mysql_select_db("arts_shows") or die(mysql_error()); echo "<h2>Search Results for $find in $field</h2>\n"; $find = addslashes($find); $result = mysql_query("SELECT * FROM craft_shows WHERE $field LIKE '%$find%'"); if(mysql_num_rows($result) == 0) { echo "<p>0 matches found.</p>"; } else { echo "<table><tr><td class=\"shows\">"; $sql = "SELECT *, DATE_FORMAT(`start_date`, '%b %e, %Y') AS s_date FROM craft_shows"; $result = mysql_query($sql); while($row = mysql_fetch_array($result)) { echo "<a href='/show_submits/show_detail.php?id={$row['id']}'>Details</a>\n"; echo $row['venue_state'] . " {$row['s_date']} {$row['show_name']} {$row['venue_city']}<br>\n"; } echo "</td></tr></table>\n"; } } ?> thanks Similar TutorialsThis code works when theres no predefined value however I'm trying to modify it in case there IS a value from the DB which is from the " . $row2['content'] . " data. Code: [Select] echo "<select name=" . $row2['ID'] . " id=" . $row2['ID'] . " class=dropdown biofield title=" . $row2['fullName'] . " >"; echo "<option value= >None</option>"; if ($styleID == 1 || $styleID == 2 || $styleID == 6) { $charactersQuery = " SELECT characters.ID, characters.characterName FROM characters WHERE characters.styleID = 3 ORDER BY characters.characterName"; } else { $charactersQuery = " SELECT characters.ID, characters.characterName FROM characters WHERE characters.styleID IN (1,2,6) ORDER BY characters.characterName"; } $charactersResult = mysqli_query ( $dbc, $charactersQuery ); // Run The Query while ( $row3 = mysqli_fetch_array ($charactersResult, MYSQLI_ASSOC)) { echo "<option value=" . $row3['ID'] . ">" . $row3['characterName'] . "</option>\r"; } echo "</select>"; I normally use something like this but with the other code not sure how to modify it to work the same way. Code: [Select] while ( $champion_row = mysqli_fetch_array ( $charactersResult, MYSQL_ASSOC ) ) { print "<option value=\"".$champion_row['ID']."\" "; if($champion_row['ID'] == $row['champID']) { print " SELECTED"; } print ">".$champion_row['characterName']."</option>\r"; } Thank you for writing this script. It has been most helpful in doing a search for a website I manage. Everything is working fine but I would like some help in modifying the "$results[] = " line. For instance, I would like to show or hide the address of a business if a data value is set to 1 if set to 0 then hide. Same for the business web address. if ( $row_memberrs['WebAddress'] != NULL ) { echo <a href=\"{$row['WebAddress']}\" target=_blank>Visit our website</a>; } I am a newbie to php and would appreciate any help you can give me. Here is a link to the search page http://www.wildwoodba.org/searchsite.php If it is not too much trouble I would like to hide the check boxes and make it search the body, title or disc for the words entered. Thank you for your help. I am currently trying to build a contact form for a website. I have been given a script to use that was used on another site but it has an iff statement in it to change the strEmail according to the location and requirement. My first real attempt to mess with MySql. Keep getting a syntax error and I am quite confused. Error: Code: [Select] Parse error: syntax error, unexpected '}' in /home/a8152576/public_html/MemberSystem/install1.php on line 68 Here is my php code: Code: [Select] <!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>Untitled Document</title> </head> <body> <form id="form1" name="form1" method="post" action=""> <h2><center>Fill in the following info properly to install required databases.</center></h2> <p> <label>Host: <input type="text" name="host" id="host" /> </label> </p> <p> <label>Database User: <input type="text" name="dbUser" id="dbUser" /> </label> </p> <p> <label>Database Password: <input type="text" name="dbPass" id="dbPass" /> </label> </p> <p> <label>Desired Admin Username: <input type="text" name="adminName" id="adminName" /> </label> </p> <p> <label>Desired Admin Password: <input type="text" name="adminPass" id="adminPass" /> </label> </p> <p> <label>Email: <input type="text" name="adminEmail" id="adminEmail" /> </label> </p> <p> <input type="submit" name="submitBtn" id="submitBtn" value="Submit" /> </p> </form> <?php if(isset($_POST['submit'])){ $hostName = $_POST['host']; $dbUserName = $_POST['dbUser']; $dbPassword = $_POST['dbPass']; $adminUserName = $_POST['adminName']; $adminPassword = $_POST['adminPass']; $email = $_POST['adminEmail']; $sqlLink = mysql_connect('$hostName','$dbUserName','$dbPassword'); if(!$sqlLink){ die('Could not connect: ' .mysql_error()); mysql_close(); } else { $queryCreate = 'CREATE DATABASE member_db'; echo 'Connected successfully!'; if(mysql_query($queryCreate,$link)){ echo 'Created database!'; mysql_select_db('member_db'); $createAdminTable = "CREATE TABLE " .$adminUserName. " (`secLvl` tinyint(1) default NOT NULL,`rank` tinyint(2) default NOT NULL,`username` varChar(32) NOT NULL UNIQUE,`password` varChar(32) NOT NULL,`email` varChar(32) NOT NULL UNIQUE"; $insertAdminInfoQuery = "UPDATE " .$adminUserName. " SET `secLvl` = '5', `rank` = '10', `username` = '$adminUserName', `password` = '$adminPassword', `email` = '$email' WHERE '$adminUserName' LIMIT 1"; mysql_query($createAdminTable) or die("ERROR: " .mysql_error()); mysql_query($insertAdminInfoQuery) or die("ERROR: " .mysql_error()) } else { echo 'Error: '.mysql_error(); } } } ?> </body> </html> You can even see the error on this web page directly: http://www.ptcptc.info/MemberSystem/install1.php Thanks, Brandon This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=355561.0 This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=327718.0 I have a table in my database which stores lot of file names & file path. iam currently listing the files on the basis of its uploaded date (stored in databse) but I want to list it on the basis of its file modification time. How i can do it ?? my current sql statement is Code: [Select] $sql = mysql_query("SELECT * FROM `files` WHERE `cid` = '$types' ORDER BY `uploaded` DESC LIMIT $start,$limit"); anyone can help me ?? Hi all. I have the following function which auto-generates a form based on a database table. What I want to do is "cancel" the function if the table doesn't exist. The function as is... function build_form($table_name){ $sql="SELECT * FROM $table_name"; $result=mysql_query($sql); $num=mysql_num_rows($result); $i=0; echo "<form method=\"post\" action=\"/php/process_data.php\">"; echo "<input type=\"hidden\" name=\"selected_table\" value=\"" . $table_name . "\"/>"; echo "<table>"; echo "<tr><td colspan=\"2\" style=\"font:1em arial;font-weight:bold;text-align:center;\">Input Form: " . $table_name ."</td></tr>"; $field_names=array(); while ($i < mysql_num_fields($result)){ $fields=mysql_fetch_field($result,$i); echo "<tr><td>" . $fields->name . "</td><td><input type=\"text\" size=\"30\" name=\"" . $fields->name . "\" /></td></tr>"; $i++; }; echo "<tr><td colspan=\"2\" style=\"text-align:center;\"><input type=\"submit\" value=\"Submit Data\" style=\"width:75%\" /></td></tr>"; echo "</table>"; echo "</form>"; }; Would something like this work? (note lines 3-5) function build_form($table_name){ $sql="SELECT * FROM $table_name"; if(!$sql){ return; }; $result=mysql_query($sql); $num=mysql_num_rows($result); $i=0; echo "<form method=\"post\" action=\"/php/process_data.php\">"; echo "<input type=\"hidden\" name=\"selected_table\" value=\"" . $table_name . "\"/>"; echo "<table>"; echo "<tr><td colspan=\"2\" style=\"font:1em arial;font-weight:bold;text-align:center;\">Input Form: " . $table_name ."</td></tr>"; $field_names=array(); while ($i < mysql_num_fields($result)){ $fields=mysql_fetch_field($result,$i); echo "<tr><td>" . $fields->name . "</td><td><input type=\"text\" size=\"30\" name=\"" . $fields->name . "\" /></td></tr>"; $i++; }; echo "<tr><td colspan=\"2\" style=\"text-align:center;\"><input type=\"submit\" value=\"Submit Data\" style=\"width:75%\" /></td></tr>"; echo "</table>"; echo "</form>"; }; Hi,
I have the following code which gives a fading out effect when clicking on links. I need to be able to disable it for anchor links so it is disabled when using # as the link.
Any ideas ?
/*
* Function to animate leaving a page
*/
$.fn.leavePage = function() {
this.click(function(event){
// Don't go to the next page yet.
event.preventDefault();
linkLocation = this.href;
// Fade out this page first.
$('body').fadeOut(400, function(){
// Then go to the next page.
window.location = linkLocation;
});
});
};
Thanks,
Scott.
Hello Coder Guys, I need a small help from you. I want to list all files & directories based on last file modification time of server. I want to list new files as first & old files as last. Consider "files" as the directory name which contains all files & directories. Please make the php code to display all files & directories from "files" directory based on last file modification time of server. Thanks IN Advance ! This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=332575.0 Hello Freaks! I am having problem with following code: <?php simplexml_load_string ('?xml version="1.0" encoding="utf-8"?>'); php?> This code is giving me following error Parse error: syntax error, unexpected '<' in public_html/wp-content/plugins/pluginnamehere/whatever.php on line 665 I know there is only a minor bracket problem that causing me this error. I would be helpful to you if you can modify it for me Thanks Hey, I'm using a script which allows you to click on a calendar to select the date to submit to the database. The date is submitted like this: 2014-02-08 Is there a really simple way to prevent rows showing if the date is in the past? Something like this: if($currentdate < 2014-02-08 || $currentdate == 2014-02-08) { } Thanks very much, Jack Hi Guys...i have this problem... I have this page where user can type in their date of birth...I would like to convert the date of birth input by the user to the DATA format which is defined by phpmyadmin (YYY-MM-DD) so that i can put into the database... My main objective of this is to calculate age... I have this form to allow use to type date of birth: Code: [Select] <label for="dob">D.O.B.: </label> <input type="text" name="dob" id="dob" class="regfields"/> This is to store the input to a variable: Code: [Select] $adddob = $_POST['dob']; and finally this query to insert all the input into the database: Code: [Select] $query = "insert into emp (FNAME, LNAME, CDSID, PAYNO, MAIL , TELNO , DOB , BRANCH , LICNO , CLASS , VFROM , VTO , EID , PASS , PIN) values ('$addfname','$addlname','$addcdsid','$addpayno','$addmail','$addtelno','$adddob','$addbranch','$addlicno','$addclass','$addvfrom','$addvto','$addeid','$addpassword','$addpin')"; Thanks in advance...If this is possible then I will set my dob column to the DATA format... Hi, I have the date stored in my datrabase like this: 2010-04-03 I am then outputting the date like this: Sat 13 Nov This is done using this code: list($year,$month,$day) = explode('-',$Coursedate); $time_stamp = mktime(0,0,0,$month,$day); $Coursedate2 = date("D d M",$time_stamp); The problem I have come across is when listing a date for next year. The database has a value of 2011-01-29 which is a Saturday but it is outputting it as a Friday: Fri 29 Jan The 29th of January 2010 is a Friday but 2010 is a Saturday. Any idea what is wrong with the code? Code: [Select] <?php $dayNames = array("Nedelja", "Ponedeljak", "Utorak", "Srijeda", "Četvrtak", "Petak", "Subota"); $dan = $dayNames[date('N')]; echo $dan.", ".date('d.m.Y.') ; ?> This is my code for showing days in my language, but it works on every day except sunday! Does anybody have any clue? Thanks! hy every one i have two files one is index.php and second insert.php as follow, index.php Code: [Select] <html> <body> <form action="insert.php" method="post"> Time <input type="text" name="time" /> <input type="submit" /> </form> <?php $ctime = date('h:i:s A'); echo "Current time is : " . $ctime ; echo "<br>"; $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("time", $con); $result = mysql_query("SELECT * FROM time"); while($row = mysql_fetch_array($result)) { echo date('h:i:s A', strtotime($row['time'])); echo "<br />"; } mysql_close($con); ?> </body> </html> here is my insert.php file code, Code: [Select] <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("time", $con); $sql="INSERT INTO time (time ) VALUES ('$_POST[time]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; echo "<br>"; echo "<a href='index.php'>Home</a>"; mysql_close($con) ?> this code work fine but the form which submit the value of time in database whos code is in index.php file above, i have to enter time in the form text field like this 17:00 and it save the value in database but, i want to enter time in the form text field like this 05:00 PM which automatically save the value in database as 17:00 what should i do with the code help me please thanks Hi, I'm preparing a code for an automated email, but the date time is not working correctly. I wanted to do a cron scheduler that emails people twice a day - 9am and 9pm. 9am sends email about records logged starting 9pm yesterday to 9am, while 9pm sends email about records logged from 9:01am to 9:00pm. The database uses date/time, so I was doing an if statement of date time, but I wasn't getting the result I wanted. Sometimes the $now time is not within the yesterday 9pm to the 9pm today range. I don't know what I'm doing wrong, but I'm sure that it has something to do with the if statement, as the some of the $now time enters the else statement Here is my code: <?php $today = date('Y-m-d 00:00:00'); $nineyesterday = date('Y-m-d H:i:s', mktime(date("H") - (date("H") + 6), date("i") - date("i"), date("s") - date("s"), date("m") , date("d"), date("Y"))); $now = date('Y-m-d H:i:s', mktime(date("H") + 13, date("i"), date("s"), date("m") , date("d"), date("Y"))); $nineam = date('Y-m-d 09:00:00'); $nine30 = date('Y-m-d 09:30:00'); $ninepm = date('Y-m-d 18:00:00'); echo '9pm yesterday: '.$nineyesterday; echo '<br/>'; echo '9am: '.$nineam; echo '<br/>'; echo '9pm: '.$ninepm; echo '<br/>'; echo 'now: '.$now; echo '<br/>'; if ((strtotime($now) > strtotime($nineyesterday)) and (strtotime($now) <= strtotime($nineam))) { //count the answers by yes or no echo 'am'; } elseif ((strtotime($now) > strtotime($nineam)) and (strtotime($now) <= strtotime($ninepm))) { //count the answers by yes or no echo 'pm'; } else { echo 'error'; } ?> Thanks in advance I have such a little code. If I echo $order_date, I receive normal date in YYYY-MM-DD format. But when I use $order_date in mysql query it is not showing me anything. I do not receive any error or warning also. Please help me. $query = "SELECT Product,Amount FROM orders WHERE Date='$order_date'"; $result = mysql_query($query) or die(mysql_error()); while($products_list = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $products_list["Product"] . "</td>"; echo "<td>" . $products_list["Amount"] . "</td>"; echo "</tr>"; } |
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