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PHP - Echo Day In Dropdown
I cannot get this to echo out the day from mysql. Basically I need it to say if the field is empty show Day as selected=selected with the list of days, but if it isn't empty, make the day in mysql set to selected=selected.
I have also check the query in mysql so I know the query is correct. <?php if (empty($r['dayofbirth'])) { $isset = isset($_POST['date_of_birth']); for ($day = 1; $day <= 31; ++$day) { echo '<option'; if ($isset && $_POST['date_of_birth'] === $day) { echo 'selected="selected"'; } echo ">${day}</option>"; } else { echo "${r['dayofbirth']}"; } ?> Similar TutorialsI'm trying to do a couple of things using the dropdown list below. I have a table with id, rank, and amount. 1. I need to echo the amount that corresponds to the rank selected from the dropdown list. 2. I also need to put the amount value in a variable so I can do some math with it later on. I've got the dropdown pulling the rank values from the db just not sure how to make it do the rest I'm trying to do. Any help would be greatly appreciated Code: [Select] <?php db($host,$db_name,$user,$pass); function db($host,$db_name,$user,$pass) { global $link; $link=mysql_connect ("$host","$user","$pass"); if(!$link){die("Could not connect to MySQL");} mysql_select_db("$db_name",$link) or die ("could not open db".mysql_error()); } $sql="SELECT * FROM dlaw"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $rank=$row["rank"]; $options.="<OPTION VALUE=\"$rank\">".$rank.'</option>'; } ?> <SELECT NAME=rank><OPTION VALUE=0>Select Rank<?=$options?></SELECT> the line : echo $form['catcher_id'] gives me a dropdown list when i choose another item from the dropdown i want to do a few things but my code not working: $selected_catcher = $form['catcher_id']; foreach($selected_catcher as $val) { $catcher_name = $val->getName(); echo $catcher_name." ".$val->getId(); if ($catcher_name = "zed-catcher") { echo $form['service_code']->renderLabel(); echo $form['service_code']->renderError(); echo $form['service_code']; } } please help? thanks OK, have no idea what's going on... I've done this a million times... why wont this output!?? I must have a major brain meltdown and dont know it yet!!! Code: [Select] <?php // this echoes just fine: echo $_POST['testfield']; // but this wont echo: echo if (isset($_POST['testfield'])) { $_POST['testfield'] = $test; } echo $test; /// or even this DOESNT echo either!: $_POST['testfield'] = $test; echo $test; ?> Hi All, I'm trying to echo the response from an SLA query, the query works and returns the data when I test it on an SQL application.. but when I run it on my webpage it won't echo the result. Please help? <?php $mysqli = mysqli_connect("removed", "removed", "removed", "removed"); $sql = "SELECT posts.message FROM posts INNER JOIN threads ON posts.pid=threads.firstpost WHERE threads.firstpost='1'"; $result = mysqli_query($mysqli, $sql); echo {$result['message']}; ?> I have a form with dropdown list that is populated with values from Sql Server table. Now i would like to use this selected item in SQL query. The results of this query should be shown in label or text field. So when a user selects item from dropdown menu, results from SQL query are shown at the same time. I have two dropdown list at the moment in my form. First one gets all values from a column in table in SQL Server. And the second one should get a value from the same table based on a selection in first dropdown list.
When i load ajax.php i get 2 error mesages: This is my code so far. I have tried to do it with this Ajax script. But i can only get first dropdown to work. The second dropdown(sub_machinery) does not show values, when first dropdown item is selected. The second dropdown should show values from databse table with this query( *$machineryID* is first dropdown selected item): SELECT MachineID FROM T013 WHERE Machinery=".$machineryID. Index.php
<!doctype html>
<?PHP
$server = "server";
$options = array( "UID" => "user", "PWD" => "pass", "Database" =>
"database");
$conn2 = sqlsrv_connect($server, $options);
if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
echo " ";
?>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>
<body>
<section id="formaT2" class="formaT2 formContent">
<div class="row">
<div class="col-md-2 col-3 row-color remove-mob"></div>
<div class="col-md-5 col-9 bg-img" style="padding-left: 0;
padding-right: 0;">
<h1>Form</h1>
<div class="rest-text">
<div class="contactFrm">
<p class="statusMsg <?php echo
!empty($msgClass)?$msgClass:''; ?>"><?php echo $statusMsg; ?></p>
<form action="connection.php" method="post">
<div>machinery</div>
<select id="machinery">
<option value="0">--Please Select Machinery--</option>
<?php
// Fetch Department
$sql = "SELECT Machinery FROM T013";
$machanery_data = sqlsrv_query($conn2,$sql);
while($row = sqlsrv_fetch_array($machanery_data) ){
$id = $row['Id'];
$machinery = $row['Machinery'];
// Option
echo "<option value='".$id."' >".$machinery."</option>";
}
?>
</select>
<div class="clear"></div>
<div>Sub Machinery</div>
<select id="sub_machinery">
<option value="0">- Select -</option>
</select>
<input type="submit" name="submit"
id="submit" class="strelka-send" value="Insert">
<div class="clear"> </div>
</form>
</div>
</div>
</div>
</div>
</section>
</script>
<script type="text/javascript">
$(document).ready(function(){
$("#machinery").change(function(){
var machinery_id = $(this).val();
$.ajax({
url:'ajaxfile.php',
type: 'post',
data: {machinery:machinery_id},
dataType: 'json',
success:function(response){
var len = response.length;
$("#sub_machinery").empty();
for( var i = 0; i<len; i++){
var machinery_id = response[i]['machinery_id'];
var machinery = response[i]['machinery'];
$("#sub_machinery").append("<option
value='"+machinery_id+"'>"+machinery+"</option>");
}
}
});
});
});
</script>
</body>
</html>
Ajaxfile.php
<?php
$server = "server";
$options = array( "UID" => "user", "PWD" => "pass",
"Database" => "database");
$conn2 = sqlsrv_connect($server, $options);
if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
echo " ";
$machineryID = $_POST['machinery']; // department id
$sql = "SELECT MachineID FROM T013 WHERE Machinery=".$machineryID;
$result = sqlsrv_query($conn2,$sql);
$machinery_arr = array();
while( $row = sqlsrv_fetch_array($result) ){
$machinery_id = $row['ID'];
$machinery = $row['MachineID'];
$machinery_arr[] = array("ID" => $machinery_id, "MachineID" =>
$machinery);
}
// encoding array to json format
echo json_encode($machinery_arr);
?>
Edited May 6, 2019 by davidd
Hi freaks, I'm new to php first of all. I'm dynamically binding a dropdownlist with mysql database . After the user selects an item from it , I want to match that item with another table so as to populate another database. The code I'm using to populate dropdown: Code: [Select] <?php $con = mysql_connect("localhost","root",""); if(!$con) { die ('Can not connect to : '.mysql_error()); } mysql_select_db("ims",$con); $result=mysql_query("select cat_id,cat_name from category"); echo "<select name=cat>"; while($nt=mysql_fetch_array($result)) { echo "<option value=$nt[cat_id]> $nt[cat_name] </option>"; } echo "</select>"; mysql_close($con); ?> Now after the user selects any one of the item , I want to bind another dropdown on the same page using such query like $result=mysql_query("select subcategory.sc_id,subactegory.sc_name from subcategory,category where subcategory.sc_id=$nt[cat_id]"); Please anyone tell me the logic and code to do it. Also tell me do I need an intermediate page to post the 1st dropdown value and then continue with 2nd dropdown. I couldn't figure out the concept anyhow. Help on this will be highly appreiable . (Tell me if I'm not clear with my question) So I need to echo a row from my database with php, but where i need to echo is already inside an echo. This is my part of my code: $con = mysql_connect("$host","$username","$password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("main", $con); $result = mysql_query("SELECT * FROM Vendor"); while($row = mysql_fetch_array($result)) { //I need to echo right here .................. but I get a blank page when I try this. Please Help. echo '<option value=$row['vendor_id']>'; echo $row['vendor_id']; echo '</option>'; } mysql_close($con); Result: A Blank page. Thanks in advance! I have a log system that allows 10 logs on each side(Left and right). I am trying to make it so that the left side has the 10 most recent logs, then the right as the next 10. Any ideas? Hi
I try to echo out random lines of a html file and want after submit password to whole content of the same html file. I have two Problems.
1st Problem When I echo out the random lines of the html file I don't get just the text but the code of the html file as well. I don't want that. I just want the text. How to do that?
for($x = 1;$x<=40;$x++) {
$lines = file("$filename.html");
echo $lines[rand(0, count($lines)-1)]."<br>";
}
I tried instead of "file("$filename.html");" "readfile("$filename.html");" But then I get the random lines plus the whole content. Is there anything else I can use instead of file so that I get the random lines of text without the html code?P.S file_get_contents doesn't work either have tried that one.
2nd Problem:
As you could see in my first problem I have a file called $filename.html. After I submit the value of a password I want the whole content. But it is like the program did forget what $filename.html is. How can I make the program remember what $filename.html is? Or with other words how to get the whole content of the html file?
My code:
if($_POST['submitPasswordIT']){
if ($_POST['passIT']== $password ){
$my_file = file_get_contents("$filename.html");
echo $my_file;
}
else{
echo "You entered wrong password";
}
}
If the password isn't correct I get: You entered wrong password. If the password is correct I get nothing. I probably need to create a path to the file "$filename.html", but I don't know exactly how to do that.
I am not a developer but I can modify code to work for me. The following code works on my test machine (Windows 10, IIS, PHP 7.4) but doesn't work on my website (cPanel, Some version of Linux, PHP 7.4). The two dropdowns are for State and City. You are supposed to be able to select the state and then select a city from that state then bring up a report for craft breweries in the city. When selecting State from the first dropdown, the page refreshes, the URL is correct with the reports.php?cat=<STATE> so $cat is being set, but the first dropdown no longer has the state selected and the second dropdown is populated with All cities and not just one ones from the selected state. Any ides why this is working fine on one machine and not the other? Selected code from reports.php
<?php
?>
/////// for second drop down list we will check if State is selected else we will display all the cities/////
echo "<form method=post action='brewerylistbycity.php'>";
}
////////// Starting of second drop downlist /////////
//// End Form /////
Hi all, I have a page which simply pulls info from a database by id: <?php include ('connect.php'); $id = $_GET['id']; $query = mysql_query("SELECT * FROM JOBS WHERE id=$id"); if (!$query) { echo "Could not run query: " . mysql_error(); exit; } $row = mysql_fetch_row($query); { echo "<body><h3>" . $row[1]. "</h3>"; echo "<h4>" . $row[2] . "</h4>"; echo "<h4>" . $row[3] . "</h4>"; echo "<h5>Duties & Responsibilities:</h5><ul>"; echo "<li><strong>" . $row[4] . "</strong>" . $row[5] . "</li>"; echo "<li><strong>" . $row[6] . "</strong>" . $row[7] . "</li>"; echo "<li><strong>" . $row[8] . "</strong>" . $row[9] . "</li>"; echo "<li><strong>" . $row[10] . "</strong>" . $row[11] . "</li>"; } ?> However in some cases the rows in the database may only contain data upto row 6 for example, how would I go about coding this so that it only displays rows that exist. If row 6 exists then 7 always will too as the information is connected. I am manually added this stuff into phpmyadmin as I do not need a form as once it is complete then it will not need to be added to. Also row 8 and 9 may contain data but 6 and 7 may not Many Thanks I want to echo this div below but only if there is more than one of these fields missing. How would I go about this? <?php //only echo all of this if one or more of these fields does not exists so....... if () { echo" <div id=\"items_todo\"> <h3>Members Info</h3><br/> <ul>"; if (empty($User['address_L1'])) { echo "<li><a href=\"#\">Update Address</a></li>";} elseif (empty($User['intro'])) {echo "<li><a href=\"#\">About You</a></li>";} elseif (empty($User['profile_image'])){echo "<li><a href=\"#\">Add Profile Image</a></li>";} elseif (empty($User['Nickname'])){echo "<li><a href=\"#\">Add A Nickname</a></li>";} echo"</ul> </div>"; } ?> Hi all I am trying to get the contents from an array using the below SQL query: $sql = mysql_query ("SELECT MAX (postage_world) FROM `basket` "); while ($row = mysql_fetch_array($sql)) { echo $row["id"]; echo $row["description"]; echo $row["postage_world"]; } I need the SQL query to find the highest value from the table and then echo out the results. The error I get is "Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given" Many thanks for your help! How do I echo the day from the db? It just displays day and I want it to display the day that was set. <select name="date_of_birth" id="date_of_birth"> <option value="">Day</option> <?php $isset = isset($_POST['date_of_birth']); for($day = 1; $day <= 31; ++$day) { echo '<option'; if ($isset && $_POST['date_of_birth'] === $day) { echo ' selected="selected"'; } echo ">${day}</option>\n\t\t\t\t\t\t"; } ?> </select> What is the correct way to write an if/else statement within an echo? I need to alter this so that I can query to see what data is found and if not correct not to echo the rest of the statement. Code: [Select] echo '<td class="productbox"> <h1>' . $product_title . '</h1> </td>'; So from the above code which is echoed within the single quotes, what is the correct way to include an if else check on a value from the database. I know how its done, but just want to save time and write it the correct way within this. Hi I once found out how to do this 4 years ago but I lost my backups and memory on the matter. I want this xml to be echoed? out as it is. When I try the standard echo and print, I get the first two lines of the code in gray when I view the page's source. I get no xml at all. Code: [Select] <?xml version="1.0" encoding="utf-8"?> <!DOCTYPE blah [ <!ELEMENT cart (title, items)> <!ELEMENT title (#PCDATA)> <!ELEMENT items (item)+> <!ELEMENT item (prive, deprive, onprive+)> <!ELEMENT security (#PCDATA)> <!ELEMENT answer (#PCDATA)> <!ATTLIST answer correct (yeah) #IMPLIED> ]> this is followed by the standard displaying of xml items could anyone help me over here? thanks in advance... Hi Freaks, You guys helped me yesterday & now I have expanded on what I was doing & now need more help. From my limited knowledge I'm trying to add theses strings together. (I'm a printer) Its all adding up correctly until the end. *$row['markup'] is only multiplying the last string and not the complete echo...I have indicated the last string in blue. <?php echo ($row['laminating']*$row['qty'])+($row['Stock1']*$row['qty'])+($row['cutting']+$row['production']+$row['additional']+$row['pluscover']+$row['selfcover'])*$row['markup'];?> What I need to do is add up everything in red and * by my markup value (12%) <?php echo ($row['laminating']*$row['qty'])+($row['Stock1']*$row['qty'])+($row['cutting']+$row['production']+$row['additional']+$row['pluscover']+$row['selfcover'])*$row['markup'];?> any help would be appreciated! cuzzmunger I saw that <?=$var?> instead of <?php echo $var; ?> can be used for echo, does it work on every server? |
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