|
PHP - Select Where Date Column - Today?
is it possible to do something like
Code: [Select] $today = date("Y-m-d"); $result = mysql_query("SELECT * FROM staff where date = '.$today.' "); also, is it normal for the first entry in the database not to be displayed? i have 6 entries in a table and only 2-6 are shown. when i changed the id for 1 to 7, it only displayed 3-7. Similar TutorialsOk basic setup is Table name = unit_data I have a field unit_paid_date colum type = DATE unit_paid_date = 2011-03-02 thats yyyy-mm-dd lets say for example the DueDate is the Second of every month I need to select all records from unit_data that are PAST DUE so if the last paid date is 2011-01-02 that record will pop up as PAST DUE BUT if that paid date is say 2011-03-10 it will not be shown because the invoice was paid Ahead of due date I had this almost working properly -- but can use any help you guys offer. I was attempting to use mysql WHERE queries and do checks against paid_date but failed Thanks Hi, I have a job listing website which displays the closing date of applications using: $expired_date (This displays a date such as 31st December 2019) I am trying to show a countdown/number of days left until the closing date. I have put this together, but I can't get it to show the number of days.
<?php
$expired_date = get_post_meta( $post->ID, '_job_expires', true );
$hide_expiration = get_post_meta( $post->ID, '_hide_expiration', true );
if(empty($hide_expiration )) {
if(!empty($expired_date)) { ?>
<span><?php echo date_i18n( get_option( 'date_format' ), strtotime( get_post_meta( $post->ID, '_job_expires', true ) ) ) ?></span>
<?php
$datetime1 = new DateTime($expired_date);
$datetime2 = date('d');
$interval = $datetime1->diff($datetime2);
echo $interval->d;
?>
<?php }
}
?>
Can anyone help me with what I have wrong? Many thanks (continuing from topic title) So if I set a date of July 7 2011 into my script, hard coded in, I would like the current date to be checked against the hard coded date, and return true if the current date is within a week leading up to the hard coded date. How could I go about doing this easily? I've been researching dates in php but I can't seem to work out the best way to achieve what I'm after. Cheers Denno Hi guys, I'm putting together a small event system where I want the user to add his own date and time into a textfield (I'll probably make this a series of drop-downs/a date picker later). This is then stored as a timestamp - "0000-00-00 00:00:00" which displays fine until I try to echo it out as a UK date in this format - jS F Y, which just gives today's date but not the inputted date. Here's the code I have right now: Code: [Select] $result = mysql_query("SELECT * FROM stuff.events ORDER BY eventdate ASC"); echo "<br />"; echo mysql_result($result, $i, 'eventvenue'); echo ", "; $dt = new DateTime($eventdate); echo $dt->format("jS F Y"); In my mysql table eventdate is set up as follows: field - eventdate type - timestamp length/values - blank default - current_timestamp collation - blank attributes - on update CURRENT_TIMESTAMP null - blank auto_increment - blank Any help as to why this could be happening would be much appreciated, thanks. Is there a way of getting today's date (in European format - day-month-year) into the body of an email sent via phpmailer? Many thanks. I'm looking for a simple little code to display today's date, month, day, year and countdown to 365 days. Can anyone please help. still shows all records in the database... any idea how i go about this? Code: [Select] $date= date('y-m-d'); $query=mysql_query("SELECT * FROM listing WHERE date >= $date") or die (mysql_error()); Hellow, i need help please, writing code and it doesn't work. please help...
Here it is
WHERE start_date BETWEEN 'start_date".strtotime('-3 day')."' AND 'start_date'";
without this code everithing works fine
Thank you
I have a calendar select date function for my form that returns the date in the calendar format for USA: 02/16/2012. I need to have this appear as is for the form and in the db for the 'record_date' column, but I need to format this date in mysql DATE format (2012-02-16) and submit it at the same time with another column name 'new_date' in the database in a hidden input field. Is there a way to do this possibly with a temporary table or something? Any ideas would be welcome. Doug The Script:
<?php
include("connect.php");
?>
<?php
echo "<h1>The Hashtag: </h1>";
if(isset($_GET['hashtag'])){
echo $_GET['hashtag'];
}
// Select the ID of the given hashtag from the "hashtags" table.
$tqs = "SELECT `id` FROM `hashtags` WHERE `hashtag` = '" . $_GET['hashtag'] . "'";
$tqr = mysqli_query($dbc, $tqs) or die(mysqli_error($dbc));
$row = mysqli_fetch_assoc($tqr);
// This prints e.g.:
// 100
echo "<br/><br/>";
print_r($row['id']);
// Select the threads from the "thread" table which contains the hashtag ID number.
$tqs_two = "SELECT * FROM `thread` WHERE `hashtag_id` IN ('" . $row['id'] . "')";
$tqr_two = mysqli_query($dbc, $tqs_two) or die(mysqli_error($dbc));
$row_two = mysqli_fetch_assoc($tqr_two);
echo "<br/><br/>";
print_r($row_two);
?>
The script should select the rows by that ID number of the hashtag. It should look in the "hashtag_id" column of the table and see if that ID number can be found there, if it can be found there, then it should select that row.
The ID numbers are inside that "hashtag_id" column separated by commas.
Example:
98, 99, 100My Question: How to do the SQL query so it selects the rows by the hashtag ID number? Edited by glassfish, 17 October 2014 - 10:35 AM. Please I want get data from mysql; 1. The last 24 hours. 2 the last 7 days 3. the last Months 4. Last 12 hours please try and help me I have searched this forum as well as over 200 other forums and have not found the answer that is specific to my question. I have shortened my code drastically to assist in resolving this quickly -
I have a search form that has criteria for the search criteria with "virtual" "columns" in an array but it's not working. If I search one column at a time it works just fine but when I try to search 8 columns with one select I get the following error: SELECT Error: Unknown column 'achievements' in 'where clause'.
When a user selects search in Achievements, I need it to look at all 8 columns that are associated with achievements and bring back the results that match - the same as if the user selects search in Associations, I need it to look at all 5 columns and bring back the results that match.
My shortened code is as follows:
<!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>Search</title>
</head>
<body>
<form name="search" action="" method="POST">
<p>Search:</p>
<p>
Achievements/Associations: <input type="text" name="find1" /> in
<Select NAME="field1">
<Option VALUE="achievements">Achievements</option>
<Option VALUE="associations">Associations</option>
</Select>
<br><br>
Secondary Education: <input type="text" name="find2" /> in
<Select NAME="field2">
<Option VALUE="edu1sectype">Highest Certificate Attained</option>
<Option VALUE="edu1secname">Highest Grade Passed</option>
<Option VALUE="edu1secinst">Name of High School</option>
<Option VALUE="edu1secdate">Date Completed</option>
<Option VALUE="edu1secinsttyp">Type of Institution</option>
<Option VALUE="subjects">Subjects</option>
</Select>
<br><br>
<input type="hidden" name="searching" value="yes" />
<input type="submit" name="search" value="Search" />
</p>
</form>
<?php
$searching = $_POST['searching'];
$find1 = $_POST['find1'];
$field1 = $_POST['field1'];
$find2 = $_POST['find2'];
$field2 = $_POST['field2'];
if ($searching =="yes")
{
echo "<br><b>Searched For:</b> $find1 $find2<br>";
echo "<br><h2>Results</h2><p>";
//If they did not enter a search term we give them an error
// Otherwise we connect to our Database
include_once "connect_to_mysql.php";
mysql_select_db("table_name") or die(mysql_error());
// We preform a bit of filtering
$find = strtoupper($find);
$find = strip_tags($find);
$find = trim($find);
$find = mysql_real_escape_string($find);
$field = mysql_real_escape_string($field);
$data = mysql_query("SELECT * FROM table_name
WHERE upper(".$field1.") LIKE '%$find1%'
AND upper(".$field2.") LIKE '%$find2%'
") or die("SELECT Error: ".mysql_error());
$result = mysql_query("SELECT * FROM table_name
WHERE upper($field1) LIKE '%$find1%'
AND upper($field2) LIKE '%$find2%'
") or die("SELECT Error: ".mysql_error());
$num_rows = mysql_num_rows($result);
echo "There are $num_rows records:<br>";
echo '<center>';
echo "<table border='1' cellpadding='5' width='990'>";
// set table headers
echo "<tr><th>Reference</th>
<th>First Name</th>
<th>Last Name</th>
</tr>";
//get images and names in two arrays
$name= $row["name"];
$surname= $row["surname"];
$achieve1 = $row["achieve1"];
$achieve2 = $row["achieve2"];
$achieve3 = $row["achieve3"];
$achieve4 = $row["achieve4"];
$achieve5 = $row["achieve5"];
$achieve6 = $row["achieve6"];
$achieve7 = $row["achieve7"];
$achieve8 = $row["achieve8"];
$assoc1 = $row["assoc1"];
$assoc2 = $row["assoc2"];
$assoc3 = $row["assoc3"];
$assoc4 = $row["assoc4"];
$assoc5 = $row["assoc5"];
$edu1sectype = $row["edu1sectype"];
$edu1secinst = $row["edu1secinst"];
$edu1secname = $row["edu1secname"];
$edu1secdate = $row["edu1secdate"];
$edu1secinsttyp = $row["edu1secinsttyp"];
$subject1 = $row["subject1"];
$subject2 = $row["subject2"];
$subject3 = $row["subject3"];
$subject4 = $row["subject4"];
$subject5 = $row["subject5"];
$subject6 = $row["subject6"];
$subject7 = $row["subject7"];
$subject8 = $row["subject8"];
$compsoft1name = $row["compsoft1name"];
$compsoft2name = $row["compsoft2name"];
$compsoft3name = $row["compsoft3name"];
$compsoft4name = $row["compsoft4name"];
$compsoft5name = $row["compsoft5name"];
$compsoft6name = $row["compsoft6name"];
$achievements = array('achieve1', 'achieve2', 'achieve3', 'achieve4', 'achieve5', 'achieve6', 'achieve7', 'achieve8');
$associations = array('assoc1', 'assoc2', 'assoc3', 'assoc4', 'assoc5');
$subjects = array('subject1', 'subject2', 'subject3', 'subject4', 'subject5', 'subject6', 'subject7', 'subject8' );
$compsoft = array('compsoft1name', 'compsoft2name', 'compsoft3name', 'compsoft4name', 'compsoft5name', 'compsoft6name');
while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td ALIGN=LEFT>" . $row['id'] . "</td>";
echo "<td ALIGN=LEFT>" . $row['name'] . "</td>";
echo "<td ALIGN=LEFT>" . $row['surname'] . "</td>";
echo "</tr>";
}
echo "</table>";
//This counts the number or results - and if there wasn't any it gives them a little message explaining that
$anymatches=mysql_num_rows($data);
if ($anymatches == 0) {
echo "Sorry, but we can not find an entry to match your query";
}
}
?>
</body>
</html>
Any assistance will be greatly appreciated as I have been working on this website for the past 4 months which has totalled over 150 pages and this is one of the last pages left to program and it's taken 6 days to get to this search page to this point.
Hi. Maybe a tricky question? How do I reflect the content of a column from a database table in a roll down select menu in the browser? Let's say that the content of the table column is: Anna Michael These names should be reflected in this select menu like this: <select name="friends"> <option value="Choose a name">Choose a name</option> <option value="Anna">Anna</option> <option value="Michael">Michael</option> So visitors can choose a name, and thereby turn it into a variable, for reuse in the database. Best regards Morris This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=349563.0 So I have this program I purchased and it allows me to create custom form fields. I have been trying to create a date select box but have been struggling because I am just learning php. I was wondering if someone could lend a hand. I will try and give as much info as i can. First I will show a picture of the custom field box and then i will give the description of both boxes that the company gave me. Next i will give you the php code i have been trying to work with. Please if my code is not the best then i will take any advice. I found the code for the date selector online. Parsed Default Value - You may specify a variable or a function as the default value; for example $_SERVER[HTTP_USER_AGENT] or mktime() Parsed PHP Code - You can specify actual PHP code that will be used to return a variable. // You should assign the value you wish to use to the variable $str. Code: [Select] <?PHP FUNCTION buildDate($name, $m, $d, $y) { $date = DATE("m-d-Y",STRTOTIME("now")); $date_array = EXPLODE("-",$date); $now_m = ""; $now_d = ""; $now_y = ""; $month = '<select class="small" name="'.$name.'_month">'; $month .= '<option value="" selected disabled>mm</option>'; FOR ($i=1;$i<=12;$i++) { IF ( $m != "" ) { IF ( $m == $i ) { $now_m = "selected"; } }ELSEIF ( $m == "0" ) { $now_m = ""; }ELSE{ IF ( $date_array[0] == $i ) { $now_m = "selected"; } } $month .= '<option value="'.$i.'" '.$now_m.'>'.$i.'</option>'; $now_m = ""; }//END month for loop $month .= '</select>'; $day = '<select class="small" name="'.$name.'_day">'; $day .= '<option value="" selected disabled>dd</option>'; FOR ($i=1;$i<=31;$i++) { IF ( $d != "" ) { IF ( $d == $i ) { $now_d = "selected"; } }ELSEIF ( $d == "0" ) { $now_d = ""; }ELSEIF ( $d == $i ) { $now_d = "selected"; }ELSE{ IF ( $date_array[1] == $i ) { $now_d = "selected"; } } $day .= '<option value="'.$i.'" '.$now_d.'>'.$i.'</option>'; $now_d = ""; }//END day for loop $day .= '</select>'; $year = '<select class="small" name="'.$name.'_year">'; $year .= '<option value="" selected disabled>yyyy</option>'; FOR ($i=GMDATE("Y"); $i <=SUBSTR(get330Date(),0,4); $i++) { IF ( $y != "" ) { IF ( $y == $i ) { $now_y = "selected"; } }ELSEIF ( $y == "0" ) { $now_y = ""; }ELSEIF ( $y == $i ) { $now_y = "selected"; }ELSE{ IF ( $date_array[2] == $i ) { $now_y = "selected"; } } $year .= '<option value="'.$i.'" '.$now_y.'>'.$i.'</option>'; $now_y = ""; }//END year for loop $year .= '</select>'; //------------------------------------------- ECHO $month." ".$day." ".$year; }//END buildDate function // ------------------------------------ FUNCTION get330Date() { RETURN DATE("Y-m-d", MKTIME(0, 0, 0, SUBSTR(GMDATE("Y m d"),5,2), SUBSTR(GMDATE("Y m d"),8,2) + 330, SUBSTR(GMDATE("Y m d"),0,4)))."<br />"; } ?> I might just be totaly wacked here with what i am trying to do but i am sure someone will tell me one way or the other. lol hi, what is the proper way to to $now $next tuesday select * where date is between now and next tuesday thanks all hello with tis button i generate some statistics from mysql if(isset($_POST['sub1'])) { $result = mysql_query("SELECT servitoros1, COUNT(*) from history WHERE serv LIKE '%be%' group by serv1"); while($row = mysql_fetch_row($result)) { echo "<tr>"; foreach($row as $cell) echo "<td ALIGN=\"center\">$cell<FONT></td>"; echo "</tr>\n"; } mysql_free_result($result); } but each record have datetime field!! so how can i set this display betwean 2 selectable datetimes e.g. from 2/2/2011 15:45 to 3/2/2011 23:59 Hi all, I have done a script that works fine on a single basis... SELECT * FROM company WHERE so <> 0 AND so_active = 'yes' AND DATE_FORMAT(so_start_date, '%e') = '18' ORDER BY company_name ASC The problem I have is when I try to do a range... SELECT * FROM company WHERE so <> 0 AND so_active = 'yes' AND DATE_FORMAT(so_start_date, '%e') BETWEEN '5' AND '26' ORDER BY company_name ASC It should return at least 1 result as I am looking for '18', any help would be much appreciated. Many Thanks. Hi, How can I check mysql date/time field for all records with tommorows date.. ? Thanks This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=309828.0 |
|||||||