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PHP - How Do I Take Input And Put It Into A Table.
Hello people.
I am very new to PHP and have a website for microlight pilots where you can add an airfield. It is a Joomla based site and the input form can be seen here. http://microlight.co/index.php/New-Airfield I am quite prepared to re-write the whole input form again if I have to. Now here's my question. I would like the output of the form to be in an HTML table like it is on this page http://microlight.co/index.php/billaricay This is because at the moment people fill out the form, I get an e-mail with the info, and then have to put it into the airfield card manually myself. Sorry, If my explanation isn't very clear. But basically want users input into a table. Similar TutorialsHello everyone. I need help with the following PHP APP. I am running on (Version PHP 7.2.10) I am trying to have a page table form on table.php pass the input variable of “5-Numbers” to another page called table_results.php I want that variable string of “5-Numbers” to be compared against 4 arrays and output any duplicates found within each of those 4 lists. If nothing is found, I still want some visual output that reads “None found”.
Lets pretend I have the following example .. On table.php, I typed inside my table form the 5-Numbers .. INPUT: 2,15,37,13,28 On table_results.php, the 4 arrays to be compared against my input numbers “ 2,15,37,13,28” are ..
$array_A = array(2,6,8,11,14,18,24); $array_B = array(1,2,9,10,13,14,25,28,); $array_C = array(1,3,7,9,13,15,20,21,24); $array_D = array(4,5,12,22,23,27,28,29);
So my output should read as follows below .. OUTPUT:
TABLE COLUMN 1 COLUMN 2 ROW 1 Matches Found Results .. ROW 2 GROUP A: 2 ROW 3 GROUP B: 2,13,28 ROW 4 GROUP ? 13,15 ROW 5 GROUP ? 28 ROW 6 5#s Input: 2,15,37,13,28
Please let me know if anyone has any suggestions on how to go about it. Thanks. Edited January 1, 2019 by Jayfromsandiego Wanted to include image example Hi, I need some help here, the one which I have highlighted in red, 'dob' and 'gender', are not inputting any values into my database table. I'm just wondering did I miss out something important, or should I change the 'type' in my database table? Thanks <?php $dob = $_POST['dob']; $gender = $_POST['gender']; /**INSERT into tutor_profile table**/ $query2 = "INSERT INTO tutor_profile (name, nric, dob, gender, race) VALUES ('$name', '$nric', '$dob', '$gender', '$race')"; $results2 = mysqli_query($dbc, $query2) or die(mysqli_error()); ?> <html> <div> <label for="dob" class="label">Date Of Birth</label> <input name="dob" type="text" id="dob"/> </div> </html> <!--Race--> <div> <label for="race" class="label">Race</label> <?php echo '<select name="race" id="race"> <option value="">--Please select one--</option>'; $dbc = mysqli_connect('localhost', '111', '111', '111') or die(mysqli_error()); $query = ("SELECT * FROM race ORDER BY race_id ASC"); $sql = mysqli_query($dbc, $query) or die(mysqli_error()); while($data = mysqli_fetch_array($sql)) { echo'<option value="'.$data['race_id'].'">'.$data['race_name'].'</option>'; } echo '</select><br/>'; mysqli_close($dbc); ?> </div> What I'm trying to do here is generate a table based on a form in which the user selects two options. The first option tells the script which database entries to put in the table, the second option tells it how to arrange them. The first works perfectly, the second not at all - it doesn't produce an error, it just doesn't do anything. I've found a number of tutorials that seem to suggest that the problem is somewhere in my punctuation around ORDER BY '$POST[sort]' but I've been unable to find a solution that actually works. Any help would be much appreciated, my code is below. Thank you! mysql_select_db($database, $con); $result = mysql_query("SELECT * FROM main WHERE state='$_POST[state]' ORDER BY '$POST[sort]'"); if(mysql_num_rows($result)==0){ echo "<p align='left'>View Pantries by State</p><div id='stateform'> <form action='../admin/viewstate.php' method='post'> <select name='state' /> <option value='AL'>Alabama</option> <option value='AK'>Alaska</option> <option value='AZ'>Arizona</option> </select> <select name='sort' /> <option value='name'>Name</option> <option value='id'>Id</option> <option value='city'>City</option> <option value='zip'>Zip</option> <option value='timestamp'>Timestamp</option> </select> <input type='submit' value='Go'></input></form> </div>"; } else{ echo "<p align='left'>View Pantries by State</p> <div id='stateform'> <form action='../admin/viewstate.php' method='post'> <select name='state' /> <option value='AL'>Alabama</option> <option value='AK'>Alaska</option> <option value='AZ'>Arizona</option> </select> <select name='sort' /> <option value='name'>Name</option> <option value='id'>Id</option> <option value='city'>City</option> <option value='zip'>Zip</option> <option value='timestamp'>Timestamp</option> </select> <input type='submit' value='Go'></input></form> </div>"; echo "<div id='fptext'><span class='h1'>Food Pantries for " . $_POST['state'] . "</span><br><br></div>"; } echo "<table border='1' align='center' cellpadding='3' width='900px'> <tr> <th>ID</th> <th>Name</th> <th>Type</th> <th>Address</th> <th>State</th> <th>Phone</th> <th>E-mail</th> <th>Website</th> <th>Hours</th> <th>Requirements</th> <th>Additional Information</th> <th>Lat</th> <th>Lng</th> <th>Update</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['id'] . "</td>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['type'] . "</td>"; echo "<td>" . $row['address'] . "</td>"; echo "<td>" . $row['state'] . "</td>"; echo "<td>" . $row['phone'] . "</td>"; echo "<td>"; echo "<a href=mailto:".$row['email'].">".$row['email']."</a>"; echo "</td>"; echo "<td>"; echo "<a href=".$row['website']."\>".$row['website']."</a>"; echo "</td>"; echo "<td>" . $row['hours'] . "</td>"; echo "<td>" . $row['requirements'] . "</td>"; echo "<td>" . $row['additional'] . "</td>"; echo "<td>" . $row['lat'] . "</td>"; echo "<td>" . $row['lng'] . "</td>"; echo "<td><a href='../public/updatepage.php?id=".$row['id']."'>Update Pantry</a></td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Hi All,
I have a form with a jquery autocomplete input field. Once the user has entered 2 characters, there is a list of values (people names) containing this string that appears. This works good.
The values come from a database table with two fields (names and ids). The query that extracts the names also retreives the values in the id field.
What would be the syntax to set the "value" attribut of the input tag to the id of a name when the user clicks on that name in the list?
The PHP part that builds the json is :
$data[] = array( 'label' => $row['name'], 'value' => $row['name'], 'id' => $row['id'] ); echo json_encode($data);The JS begins as follows :
jQuery(document).ready(function(){
$('#input_id').autocomplete({source:'my_jquery_suggest.php', select: function(event, ui) {
$(event.target).val(ui.item.value);
$('#form_id').submit();
return false;
},
minLength:2,delay: 1000}).focus(function () {
window.pageIndex = 0;
$(this).autocomplete("search");
});
...
Thanks for your help.
I am writing a script that will parse my PHP classes and check for things like coupling, visualize my objects and connections, dependencies, check for convention usage, etc.
So, I have a simple file upload. I'm never saving the files, just get contents and dump the file and work with the string version.
I'm writing it for me, but I figure I might want to open it for others to use in the future, so I may as well write it that way to begin with -- so I need to validate user input. Problem is, the user input is supposed to be valid PHP code. I'm thinking that, as long as I'm careful, I shouldn't be executing any code contained in strings, but I'm no security expert and I want a warm fuzzy that my thought on this is correct. What kinds of things do I need to look out for? Is it possible to inject when working with strings?
My initial thought is to regex the entire file and replace key portions with known replacements. So ( and ) would become !* and !^ or $ would become @~ (combinations that -- I think -- don't make sense to php?) But that may be completely unnecessary processing time if I'm not in any danger, here. Thanks ahead of time for any help.
PS - as a side question -- what's the best way to verify a file is a php file? I know of getimagesize for images, but should I just check for <? to verify it's php? That seems like it would be too easy to fool -- then again, it might not matter much.
-Adam
I hope I can explain what is happening. I have created two forms in PHP. The first 'almost' works, i.e. it shows the data. But I have two problems - 1) the second pulldown menu is always empty and 2) $value from the first pulldown menu ALWAYS equals the last entry thus the last 'if' in the function subdomains ($domains) is always called (but still empty). The code may explain this better than me:
<!DOCTYPE html>
<html>
<body>
<!-- processDomains.php is this file - it calls itself (for testing purposes so I can see what is happening) -->
<form action="processDomains.php" method="post">
<?php
// create the domains array (there are actually several entries in the array but I cut it down for testing)
$domains = array (1 => 'Decommission', 'Migration');
echo "Select Domain:";
echo "<br>";
// Make the domain pull-down menu - this displays correctly
echo '<select name="domain">';
foreach ($domains as $key => $value) {
echo "<option value=\"$key\">$value</option>\n";
}
echo '</select>';
// input doesn't matter what is 'submitted', always goes to last $value
echo '<input type="submit" name="submit" value="Submit">';
// call function subdomains
subdomains ($value);
function subdomains ($domains) {
// define values for each array - each array contains available choices for the subdomain pulldown menu
$migration = array (1 => 'Application Migration', 'Application Patch', 'Application Upgrade');
$decommission = array (1 => 'Applications', 'Servers', 'Storage');
if ($domains === 'Migration') {
echo "Select subdomain:";
echo "<br>";
// Make the Migration pull-down menu
echo '<select name="migration">';
foreach ($migration as $key => $value) {
echo "<option value=\"$key\">$value</option>\n";
}
echo '</select>';
} else if ($domains === 'Decommission') {
/* ===
* since 'Decommission' is the last entry in the 'Domains' pulldown list, $value ALWAYS equals
* 'Decommission' and $domains equals $value. So this menu SHOULD work but is always
* empty. Thus, two problems - the pulldown menu is always empty and $value isn't based
* upon user input.
*/
echo "Select subdomain:"; // this prints so I know I'm in 'Decommission (I eliminated the echo "$domain" to show I'm always coming here)'
echo "<br>";
// Make the 'Decommission' pull-down menu
echo '<select name="decommission">';
foreach ($decommission as $key => $value) {
echo "<option value=\"$key\">$value</option>\n";
}
echo '</select>';
echo '<input type="submit" name="submit" value="Submit">'
) // end of 'if-else'
} // end of function 'subdomain'
?>
</form>
</body>
</html>
Let me say thank you in advance and I appreciate the help! I know I'm doing something (or more than one thing) wrong and I hope someone can tell me what it is.
Best Regards!
Edited by mac_gyver, 19 January 2015 - 09:37 PM. code tags around posted code please 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 Hi people, I really hope you guys can help me out today. I'm just a newbe at php and i'm having real trouble. Bassically all I want to do is have a user type in a company name in a html form. If what the user types in the form matches the company name in my php script i want the user to be sent to another page on my site. If what the user types in the form doesnt match the company name in my php script i want the user to be sent to a differnt page like an error page for example. this is my html form: Code: [Select] <form id="form1" name="form1" method="post" action="form_test.php"> <p>company name: <input type="text" name="company_name" id="company_name" /> </p> <p> <input type="submit" name="button" id="button" value="Submit" /> </p> </form> And this is the php code I'm trying to process the information on: Code: [Select] <?php $comp_name = abc; if(isset ($_POST["company_name"])){ if($_POST["company_name"] == $comp_name){ header("Location: http://www.hotmail.com"); exit(); } else{ header("Location: http://www.yahoo.com"); exit(); } } ?> The thing is i'm getting this error when i test it: Warning: Cannot modify header information - headers already sent by (output started at D:\Sites\killerphp.com\form_test.php:10) in D:\Sites\killerphp.com\form_test.php on line 17 Please can some one help me out, i'm sure this is just basic stuff but i just cant get it to work Cheers. Hi All,
I want to copy into a table values from another table that partially match a given value, case-insensitively. So far I do as follows but I wonder whether there is a quicker way.
$input_table=array('1'=>'toto','2'=>'tota','3'=>'hello','4'=>'TOTO','5'=>'toto');
$input_table_2 = array_map('strtolower', $input_table);
$value_to_look_for='Tot';
$value_to_look_for_2=strtolower($value_to_look_for);
$output_table=array();
foreach ($input_table_2 as $k=>$v)
{
if(false !== strpos($v, $value_to_look_for_2))
{
$output_table[]=$input_table[$k];
}
}
One drawback is that $input_table_2 is 'foreached' whereas there might be no occurrences, which would lead to a loss of time/resources for big arrays.
Thanks.
Hello, I need some help. Say that I have a list in my MySQL database that contains elements "A", "S", "C", "D" etc... Now, I want to generate an html table where these elements should be distributed in a random and unique way while leaving some entries of the table empty, see the picture below. But, I have no clue how to do this... Any hints? Thanks in advance, Vero Hi All ,
I have a small table with 4 fields namely Day_ID, Dues, Last_Visit, Points. where Day_ID is an auto-increment field. The table would be as follows:
Day_ID -- Dues --- Last_Visit --- Points.
1 --------- 900 -------- 1/12 -------- 6
2 --------- 700 -------- 4/12 -------- 7
3 --------- 600 -------- 7/12 -------- 5
4 --------- 600 -------- 9/12 -------- 6
5 --------- 600 -------- 10/12 ------- 6
6 --------- 600 -------- 14/12 ------- 6
So this is the record of a person's visit to say a club. The last row indicates the last date of his visit to the club. His points on this date are 6. Based on this point value of 6 in the last row I want to retrieve all the previous BUT adjoining all records that have the same Points i.e. 6.
So my query should retrieve for me, based on the column value of Points of the last row (i.e. Day_ID - 6 ), as follows:
4 --------- 600 -------- 9/12 -------- 6
5 --------- 600 -------- 10/12 ------- 6
6 --------- 600 -------- 14/12 ------- 6
This problem stated above had been completely resolved, thanks to a lot of help from Guru Barand by this following query :-
$query = "SELECT cv.day_id, cv.dues, cv.last_visit, cv.points FROM clubvisit cv WHERE last_visit >= ( SELECT MAX(last_visit) FROM clubvisit WHERE points <> ( SELECT points as lastpoints FROM clubvisit JOIN ( SELECT MAX(last_visit) as last_visit FROM clubvisit ) as latest USING (last_visit) ) )";I am using this and it works perfectly except that now there is a slight change in the table because the criteria for points is now dependent on more than one column cv.points and is more like cv.points1, cv.points2, cv.points3 etc. So now I need to make a selection based on each of these cv.points columns. As of now I can still get the results by running the query multiple times for each of the cv.points columns ( seperately for cv.points1, cv.points2, cv.points3) and it works correctly. However I am wondering if there is a better way to do this in just one go. This not only makes the code repetitive but also since the queries are interconnected, involves the use of transactions which I wish to avoid if possible. The values that I require for each of the cv.point columns is 1. day_id of the previous / old day on which the cv.points value changed from the current day value, and 2. cv.points on that old/ previous day. So for example if the table is as below: Day_ID -- Dues --- Last_Visit --- Points1 --- Points2. 1 --------- 900 -------- 1/12 ----------- 9 ------------ 5 2 --------- 600 -------- 4/12 ----------- 6 ------------ 6 3 --------- 400 -------- 7/12 ----------- 4 ------------ 7 4 --------- 500 -------- 9/12 ----------- 5 ------------ 8 5 --------- 600 -------- 10/12 ---------- 6 ------------ 8 6 --------- 600 -------- 11/12 ---------- 6 ------------ 8 7 --------- 600 -------- 13/12 ---------- 6 ------------ 7 8 --------- 500 -------- 15/12 ---------- 5 ------------ 7 9 --------- 500 -------- 19/12 ---------- 5 ------------ 7 Then I need the following set of values : 1. day_id1 -- Day 7, points1 ---- 6, days_diff1 -- (9-7 = 2) . // Difference between the latest day and day_id1 2. day_id2 -- Day 6, points2 ---- 8, days_diff2 -- (9-6 = 3) 3. day_id3 -- .... and so on for other points. Thanks all ! hi... i have a table ... i add and remove data in the table...when i add new record , information add to center of the table ! whats problem? i want add data in first of table. please guide me.thanks
Hi
I am very new to PHP & Mysql.
I am trying to insert values into two tables at the same time. One table will insert a single row and the other table will insert multiple records based on user insertion.
Everything is working well, but in my second table, 1st Table ID simply insert one time and rest of the values are inserting from 2nd table itself.
Now I want to insert the first table's ID Field value (auto-incrementing) to a specific column in the second table (only all last inserted rows).
Ripon.
Below is my Code:
<?php
$con = mysql_connect("localhost","root","aaa");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("ccc", $con);
$PI_No = $_POST['PI_No'];
$PO_No = $_POST['PO_No'];
$qry = "INSERT INTO wm_order_entry (
Order_No,
PI_No,
PO_No)
VALUES(
NULL,
'$PI_No',
'$PO_No')";
$result = @mysql_query($qry);
$val1=$_POST['Size'];
$val2=$_POST['Style'];
$val3=$_POST['Colour'];
$val4=$_POST['Season_Code'];
$val5=$_POST['Dept'];
$val6=$_POST['Sub_Item'];
$val7=$_POST['Item_Desc'];
$val8=$_POST['UPC'];
$val9=$_POST['Qty'];
$N = count($val1);
for($i=0; $i < $N; $i++)
{
$profile_query = "INSERT INTO order_entry(Size,
Style,
Colour,
Season_Code,
Dept,
Sub_Item,
Item_Desc,
UPC,
Qty,
Order_No
) VALUES( '$val1[$i]','$val2[$i]','$val3[$i]','$val4[$i]','$val5[$i]','$val6[$i]','$val7[$i]','$val8[$i]','$val9[$i]',LAST_INSERT_ID())";
$t_query=mysql_query($profile_query);
}
header("location: WMView.php");
mysql_close($con);
?>
Output is attached.Not sure how to do this at all... I'm creating a page with a form to edit existing info in two tables in the database. I need to pre-check some checkboxes and I don't know how. The first query query_selectguest below puts data into the form just fine. query_checked looks in the second table for any rows that match the discount id. Then further down in the form, I have a php snippet that pulls all the classes from a third database. As I'm echoing these out to the page, I want them to be pre-checked if they match a result found in $result_checked here's what I have right now... Code: [Select] $query_selectguest = 'SELECT * FROM tbl_discount WHERE discount_id='.$passedID; $result_guest = mysql_query($query_selectguest); $g_row = mysql_fetch_array($result_guest); $query_checked = 'SELECT * FROM active_discounts WHERE disc_id='.$passedID; $result_checked = mysql_query($query_checked); $h_row = mysql_fetch_array($result_checked); ?> <form name="form1" method="post" action="update_discount.php" onSubmit="return validate_form()"> <input name="discount_name" type="text" class="input" id="subject" size="50" maxlength="100" value="<? print $g_row['discount_name'] ?>"> <input name="discount_amount" type="text" class="input" id="subject" size="5" maxlength="3" value="<? print $g_row['discount_amount'] ?>"> <!-- here's what I'm concerned with --> This discount applies to these classes:<br> <?php $quer4=mysql_query("SELECT workshop_id, workshop_title FROM tbl_workshops order by workshop_title"); while($row4 = mysql_fetch_array($quer4)) { echo "<input type='checkbox' name='workshop_link_1[]' value='".$row4[workshop_id]."'>".$row4[workshop_title]."<BR>"; } ?> This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=317025.0 I know I'm doing it something right, but can someone tell me why only one table is showing up? Can you help me fix the issue? Heres my code: function showcoords() { echo"J3st3r's CoordVision"; $result=dbquery("SELECT alliance, region, coordx, coordy FROM ".DB_COORDFUSION.""); dbarray($result); $fields_num = mysql_num_fields($result); echo "<table border='1'>"; // printing table headers echo "<td>Alliance</td>"; echo "<td>Region</td>"; echo "<td>Coord</td>"; // printing table rows while($row = mysql_fetch_array($result)) { // $row is array... foreach( .. ) puts every element // of $row to $cell variable foreach($row AS $Cell) echo "<tr>"; echo "<td>".$row['alliance']."</td>\n"; echo "<td>".$row['region']."</td>\n"; echo "<td>".$row['coordx'].",".$row['coordy']."</td>\n"; echo "</tr>\n"; } echo "</table>"; mysql_free_result($result); } I have 2 rows inserted into my coords table. Just frustrated and ignorant to php. This is a little confusing so please bear with me. I was thinking of using left.join but couldn't figure out how to implement it properly. I am trying to order all of the results from a table names 'clans' based on how many points the clan has. To calculate the points you have to go into another table 'clanteams' and then loop every team in the clan pulling wins and losses from a row in a specific ladder table. Here is the code i have to calculate the points. Code: [Select] $total[wins] = 0; $total[loss] = 0; $members=mysql_query("SELECT id,clanid,teamid,DATE_FORMAT(datejoined,'%M %d, %Y') FROM clanteams WHERE clanid='$member[id]'"); while(list($id,$clanid,$teamid,$joined)=mysql_fetch_row($members)){ $team=mysql_query("SELECT name,ladderid FROM teams WHERE id='$teamid'"); if(mysql_num_rows($team) == 0) continue; $team=mysql_fetch_array($team); $ladder=mysql_query("SELECT name FROM ladders WHERE id='$team[ladderid]'"); $ladder=mysql_fetch_array($ladder); $linfo=mysql_query("SELECT rank,wins,loss FROM ladder_$team[ladderid] WHERE teamid='$teamid'"); $linfo=mysql_fetch_array($linfo); $total[wins] = $total[wins] + $linfo[wins]; $total[loss] = $total[loss] + $linfo[loss]; } $totalpoints = ($total[wins] * 2) - $total[loss]; So now i want to loop through every row in the clans table, and using the above code oder them by $total points. Ive spent hours wrapping my head around it and still cannot figure it out. Please help. I am new to php.i want to be able to link echoed out rows from a table to another row in another table so that when users clicked the first row i echoed out it will take them to that SPECIFIC row i linked it to in the second table..JUST LIKE FACEBOOK...Please do i make sense or is there another way to do it. Thanks in advance Simple question but I couldn't find any straight forward answers.
I have a customer details table with: customer ID, name, address, email.
I also have an delivery table with delivery ID, customer ID, order ID.
How can I link the customer ID column from the customer table to the delivery table so that when changes occur on the customer table they subsequently alter all other tables?
I know it might use foreign keys but I couldn't find any online resource that explained how to use them properly.
Cheers
How would you retrieve information from one table and enter that informatio to another and the same time get other details from form. Do you retrieve the data and use a form to post this and insert that to another table. Have I confused you??? |
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