I'm creating a save button for user to save into their database, but I having a difficulty on getting the name of the image i select.
Save button
I know using ` $image=$_GET["image"]; will get the name if the image but I dont know how to get the name outside the php because I'm using popup module.
<div id="modal01" class="w3-modal " style="padding-top:0"
onclick="this.style.display='none'">
<span class="w3-button w3-black w3-xlarge w3-display-topright">×</span>
<div class="w3-modal-content w3-animate-zoom w3-center w3-transparent w3-padding-64">
<img id="img01" class="w3-image">
<p id="caption"></p>
<form action="save.php" method="GET">
<input type="submit" name="save" value="Save"/>
</form>
</div>
Here is my code on retrieving picture from database:
<?php
while($row = mysqli_fetch_array($result)) {
echo "<div class='grid-item'><img src='images/{$row['image']}' onclick=onClick(this) style='width:98%' /></div>
";}
mysqli_close($db);
?>
Related
Why I can't insert into 1 field to my database table?
I loop the data of my other table to display each data and put the result to href tag so it clickable and concatenate with it's id so it will put the id result after the link like this:
<a href="scheduling.php?CID=<?php echo $rows['docID']; ?>">
Here is how I loop my data using php on page1:
<?php
$sel = "SELECT * FROM doctors ORDER BY docID";
$result = $conn->query($sel);
if($result->num_rows>0)
{
while($rows = $result->fetch_array())
{
?>
<a href="scheduling.php?CID=<?php echo $rows['docID']; ?>">
<div class="doc_item" style="width:310px; border:4px solid #009973;display:inline-block; margin-top:40px; margin-right:20px;">
<img src="images/ft-img.png" style="width: 300px;">
<div class="item-lvl" style="width: 100%; background:#009973; color:#fff; height:70px; padding-top:10px;">
<h4 style="font-size:20px; font-weight:bold;">Dr. <?php echo $rows['docFname']; ?></h4>
<span><?php echo $rows['docType']; ?></span>
</div>
<div style="width: 100%; background:#00664d;padding-top:15px; padding-bottom:20px;">
<h3 style="color:#fff;">Make a Schedule!</h3>
</div>
</div>
</a>
<?php
}
}
?>
Then I want to add the value of CID to my database table using this code on page 2:
<?php
if (isset($_POST['btn-submit'])) {
$cusID = $customerID;
$docID = $_GET['CID'];
$checkupType = $_POST['checkupType'];
$schedTime = $_POST['schedTime'];
$contact = $_POST['contact'];
$ins = "INSERT INTO schedule
(customerID, docID, checkupType, schedTime, contact)
VALUE ('$cusID','$docID','$checkupType','$schedTime','$contact')";
if($conn->query($ins)===TRUE) {
header('location:success_sched.php');
}else{
echo "SQL Query: " . $ins . "->Error: " . $conn->error;
}
}
Now, this code insert data to my database table except $docID which is the value is $_GET['CID'];
docID column is intiger.
please correct my code, this is my school project.
Form code on page 2:
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
<div class="sign-u">
<div class="sign-up1">
<h4>Form of Checkup* :</h4>
</div>
<div class="sign-up2">
<label>
<input type="radio" value="Virtual" name="checkupType" required>
Virtual Checkup
</label>
<label>
<input type="radio" value="Facetoface" name="checkupType" required>
Face to Face Checkup
</label>
</div>
<div class="clearfix"> </div>
</div>
<p style="color:gray;">*Take note that virtual checkup will require a you stable internet for better communication.</p>
<div class="sign-u">
<div class="sign-up1">
<h4>Date and Time* :</h4>
</div>
<div class="sign-up2">
<input type="date" name="schedTime" id="date" required>
</div>
<div class="clearfix"> </div>
</div>
<p style="color:gray;">*Please provide Skype id as a contact information if <span style="color:#ff5c33">Virtual Checkup</span> is your choice.</p>
<div class="sign-u">
<div class="sign-up1">
<h4>Contact Information :</h4>
</div>
<div class="sign-up2">
<input type="text" name="contact">
</div>
<div class="clearfix"> </div>
</div>
<div class="sub_home" style="text-align: right;">
<input style="background: #339966;" type="submit" name="btn-submit" value="Submit">
<div class="clearfix"> </div>
</div>
</form>
Please understand that when you submit a form, that is a new request to the server. It has a new URL. Your page 2 form submits to $_SERVER['PHP_SELF']; - which is the current URL but without the querystring variables. Therefore your GET parameter is not passed in the form submit. (If you use the "View Source" feature of your browser to examine the "action" attribute of the form after it's rendered, you'll see exactly what URL it makes the request to when submitting. You can also see the same thing by looking at your browser's Network tool, or your webserver's access logs.)
There are a couple of ways around that, but I'd suggest adding the value as a hidden field in the form, and populating it with the value of the GET parameter during the previous request.
e.g.
<input type="hidden" value="<?php echo $_GET["CID]; ?>" name="CID">
and then in the PHP, retrieve it via $_POST like all the other variables:
$docID = $_POST['CID'];
Finally, please ensure you update your code to use prepared statements and parameters, as recommended in the comments, so that you are protected against both SQL injection attacks and unexpected syntax errors. You can get more info and examples here
I am trying to get my jquery form to allow for multiple submissions, but it will not load after a selection.
I have a grid (let's say 2x2). I click on a cell and fill in my name from a jquery form. I click submit and my name will appear in the cell via php. However, when I go to click on another cell the pop-up window does not appear.
I have added a simplified version of my code to jsfiddle (https://jsfiddle.net/7j7wxrpu/).
You can see from there my form is a pop-up window after you click on a cell:
<table border=1>
<tr><td colspan="11"><center><h2>Away Team</h2></center></td></tr>
<tr><th class='header-cols'></th><th class='header-cols'><h1>0</h1></th><th class='header-cols'><h1>1</h1></th></tr><tr><th class='header-rows'><h1>0</h1></th><td class='grid-cells'>
<a href='#myPopup' data-rel='popup'>
<div id='cell' onclick='setCoords(0,0);'>
<div class='grid-num'>1</div>
<div class='grid-name'>justin9</div>
</div>
</a>
</td><td class='grid-cells'>
<a href='#myPopup' data-rel='popup'>
<div id='cell' onclick='setCoords(1,0);'>
<div class='grid-num'>2</div>
<div class='grid-name'>justin10</div>
</div>
</a>
</td></tr><tr><th class='header-rows'><h1>1</h1></th><td class='grid-cells'>
<a href='#myPopup' data-rel='popup'>
<div id='cell' onclick='setCoords(0,1);'>
<div class='grid-num'>3</div>
<div class='grid-name'></div>
</div>
</a>
</td><td class='grid-cells'>
<a href='#myPopup' data-rel='popup'>
<div id='cell' onclick='setCoords(1,1);'>
<div class='grid-num'>4</div>
<div class='grid-name'></div>
</div>
</a>
</td></tr></table>
<div data-role="popup" id="myPopup" class="ui-content" style="min-width:250px;">
<form method="post" action="">
<div>
<h3>Pick This Square:</h3>
<label for="name" class="ui-hidden-accessible">Name:</label>
<input type="text" name="name" id="name" placeholder="Name">
<label for="email" class="ui-hidden-accessible">Email:</label>
<input type="text" name="email" id="email" placeholder="Email">
<input type="submit" data-inline="true" value="Submit">
<!--<input type='hidden' name='row' value=''>
<input type='hidden' name='col' value=''>-->
<div id='row-div'></div>
<div id='col-div'></div>
</div>
</form>
</div>
And here is the php it calls from the file:
<?php
include_once 'connectmysql.php';
if(!isset($_POST['name']) || !isset($_POST['email'])){
//fail because one is blank
echo "Failed the POSt data: Name: " . $_POST['name'] . " | Email: " . $_POST['email'];
}
else{
$name = $_POST['name'];
$email = $_POST['email'];
$row = $_POST['row'];
$col = $_POST['col'];
$tstamp = date("Y-m-d_H:i:s");
//Write to the sql db
$conn = ConnectMySQL();
$sql = "INSERT INTO picks (name,email,paid,row,col,tstamp) VALUES('$name','$email',0,$row,$col,'$tstamp')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
}
?>
Besides the lack of security in my php is there anything I am missing? How come the pop-up box will only pop-up once until I refresh the page. I also notice when I refresh the page it tries to "Resend" the post data to the server. It looks like I have to clean the post data after a submit, is that a thing?
To do that without reloading the page you should use AJAX call to a PHP script that will insert the new data in the database and then query the database again and send the new values to your JavaScript and then with JavaScript to change the values of the cells.
Also, change names of the IDs - should have unique names: cell1, cell2.
<form id="form" name="form" method="post" action="">
<div class="jquery-script-clear"></div>
<h1>BARCODE GENERATOR</h1>
<div id="generator"> Please fill in the code :
<input type="text" name="barcodeValue" id="barcodeValue" value="1234"><br> <br>
</div>
<div id="submit">
<input type="button" onclick="generateBarcode();" value=" Generate "> <input type="button" onclick="printDiv('print')" value="Print" />
</div>
</form>
<?php
$barcodeValue = $_POST["barcodeValue"];
$save = file_get_contents("save.txt");
$save = "$barcodeValue" . $save;
file_put_contents("save.txt", $save);
echo $save;
?>
sample picture
How to save the input data in save.txt file. When i clicked generate button the text file not showing in same folder.
The problem with your code is you have no submit button so your form was not actually posting when you pressed the button. if you look at my edits you can see I changed the button from input type="button" to type="submit". That allows for the form to submit back to the same php script.
Your script also was causing errors because you accessed $_POST["barcodeValue"] without checking if it existed. You also have to check if the save.txt exists before reading from it. If analyze my edits you can see how checking if the variables are available will help quite a bit.
<form id="form" name="form" method="post" action="">
<div class="jquery-script-clear"></div>
<h1>BARCODE GENERATOR</h1>
<div id="generator"> Please fill in the code :
<input type="text" name="barcodeValue" id="barcodeValue" value="1234"><br> <br>
</div>
<div id="submit">
<input type="submit" value=" Generate ">
</div>
</form>
<?php
if(isset($_POST["barcodeValue"]))
{
$barcodeValue = $_POST["barcodeValue"];
if(file_exists("save.txt"))
$save = file_get_contents("save.txt");
else
$save = "";
$save = $barcodeValue . $save;
file_put_contents("save.txt", $save);
echo $save;
}
?>
Let me know if you need more help
I have two php page.I'm using Bootstrap, php and mysql. In the first page I load three objects into div from mysql database of the user logged. To do this I'm using the next code:
<div class="container">
<div class="row">
<?php
require_once('function.php');
conectar('localhost', 'root', '', 'mydb');
$consulta1 = mysql_query('SELECT id FROM user WHERE username="'.$_SESSION["name"].'"');
$result = mysql_query('SELECT * FROM finc WHERE Usuario_idUsuario="'.$_SESSION["idUser"].'"');
if ($row = mysql_fetch_array($result)){
do{
echo '<div class="col-lg-4">' ;
echo '<img class="center-block img-circle" src="data:image/gif;base64,R0lGODlhOw=="
alt="Generic placeholder image" style="width: 140px; height: 140px;">';
echo '<h2 class="text-center">'.$row['name'].'</h2>';
echo '<p align="center">'.$row['data'].'</p>';
echo '<p align="center">'.$row['tao'].'</p>';
echo '
<a type="button" class="btn btn-success" href="secondPage.php" role="button">Entrar »</a>
';
echo'</div>';
}while ($row = mysql_fetch_array($result));
} else {
echo "¡ No data for this user!";
<a}
?>
<!-- /.col-lg-4 -->
</div>
<!-- /.row -->
</div>
I need send the id value depending of the button clicked for load the data associated in the next php page. For example, If I click in the second button created in the do-while loop I need send the id=2 to the sencondPage.php. I have searched how to do this, but only find how to send var into url like sencondPage.php?var=2, And I hate this because user can change url... And adding value into session, but on click I haven't get this.
So, how can I pass the corresponding id value when user click in the link??
Thanks!
You could use an html <form>:
<form method="post" action="[URL FOR NEXT PAGE]">
...
<input type="submit" name="value1" value="Button 1" />
<input type="submit" name="value2" value="Button 2" />
</form>
Now if someone clicks the first button it will send them to the next page with the post data: value1=Button%201, and if they click the second button the it will instead be value2=Button%202. In either case any other form elements inside the form will also be submitted via post. With PHP you can retrieve these values using something like:
if ($_POST['value1']) {
...
elseif ($_POST['value2']) {
...
}
I have a Print button which I want to do 2 actions when clicked:
Updating my database.
Printing the HTML page.
This is what I've done so far, but it's not working:
<form action="" method="POST">
<body >
<?php
$n=$_POST['ID'];
$a=implode("</br>",$n);
list($add, $ward) = explode("(!#!)", $a);
?>
<div id="container">
<p id="address">
<?php echo"$address";?>
</p>
<p id="ward">
<?php echo"$ward";?>
</p>
</div>
<input type="submit" name="Print" value="Print" />
<?php
if(isset($_POST['Print']))
{
?><script>javascript:window.print()</script><?php
mysql_query("UPDATE `source_main` SET `source_status`=3 WHERE `source_id`=1");
}?>
<div id="footer">
</div>
</form>
After using this print button , my database is getting updated, but the print window is showing error(i.e the variables posted from other page are showing errors).
Can anyone please help me print and update at same time with this Print button?
If your page does not have it, add form tags. They are mandatory unless you want to AJAX the data. Also remove the semi-colon from the end of the sql
<form action="" method="POST">
<input type="submit" name="Print" value="Print" />
</form>
<?php
if(isset($_POST['Print'])) {
mysql_query("UPDATE `source_main` SET `source_status`=3 WHERE `source_id`=1");
?><script>window.print();</script>
<?php } ?>
UPDATE: Perhaps you mean this, but I do not want to keep correcting HTML.
<?php
$n=$_POST["ID"];
$a=implode("</br>",$n);
list($add,$ward)=explode("(!#!)", $a);
?>
<body>
<div id="headerbg">
<div id="header-e1"><a align="left" href="escalationReport.php">Back </a>
</div>
<div id="header-e3"><a align="right" href="logout.php">Logout </a>
</div>
<h1><p>Issue Notice</h1>
</div>
<div id="container">
<p id="address">
<?php echo "$address";?>
</p>
<p id="ward">
<?php echo "$ward";?>
</p>
</div>
<form action="" method="POST">
<input type="submit" name="Print" value="Print" />
</form>
<?php if(isset($_POST[ 'Print'])) {
mysql_query( "UPDATE `source_main` SET `source_status`=3 WHERE `source_id`=1");
?>
<script>
window.print();
</script>
<?php } ?>
<div id="footer"></div>
</body>
you have an error in the update query, you have placed semicolon inside the query so please remove that so the query will be like this.
Make sure your form method is POST
mysql_query("UPDATE `source_main` SET `source_status`=3 WHERE `source_id`=1");
Make sure you are using input type button inside the form tag. And use post method for form.