invalid argument supplied for foreach() php - php

i have index.php and index2.php
here is index.php:
when i submit this value, index2.php will show. Here is index2.php:
I am not understanding, why this error msg is showing.
Anyways, submitting these selected values(index2.php) it will say a disease name from the database. clicking on submit, this text is showing(below):
I am pesting my code here:
index.php
<?php
$con = mysql_connect("localhost","root","");
mysql_select_db("symptomchecker",$con) or die(mysql_error());
$q = "select dtid,diseasetype from disease_type";
$result = mysql_query($q,$con) or die(mysql_error());
$numRB = mysql_num_rows($result);
echo "<html><head><title>Disease Type</title></head><body><h1>Disease List</h1></br><form action='index2.php' method='post'>";
$i=0;
while($row = mysql_fetch_array($result)){
echo $row['diseasetype'];
echo "<input type='radio' name='diseasetype' id='pet1' value='$row[dtid]' />" ;
echo "<br />";
$i++;
}
echo " <input type='submit' value='submit' name='submit'></form></body></html>";
?>
index2.php:
$diseasetypeid = "$_POST[diseasetype]";
$j=0;$i=0;
$query = "select did from disty_dis_rel where dtid = ".$diseasetypeid."";
$result1 = mysql_query($query,$con) or die(mysql_error());
echo "<html><head><title>symptom</title></head><body><h1>symptom List</h1></br><form action='' method='post'>";
while($row = mysql_fetch_array($result1))
{
$z=$row['did'];
$query2 = "select sid,symptomname from symptom where sid = ".$z."";
$result2 = mysql_query($query2,$con) or die(mysql_error());
while($row2 = mysql_fetch_array($result2))
{
echo $row2['symptomname'];
echo "<input type='checkbox' name='pets[$i]' id='pet1' value='$row[sid]' />" ;
echo "<br />";
$i++;
}
$j++;
}
echo " <input type='submit' value='submit' name='submit'></form></body></html>";
if(isset($_REQUEST[submit]))
{
**foreach ($_REQUEST[pets] as $key=>$values)**
{
$yy.=$values."-";
}
$yy=rtrim($yy,"-");
$xx = "select did,sids from dis_sym_rel where sids=".$yy."";
$result3 = mysql_query($xx,$con) or die(mysql_error());
while($row = mysql_fetch_array($result3)){
$p=$row[did];
$xxx = "select did,diseasename from disease where did=".$p."";
$result4 = mysql_query($xxx,$con) or die(mysql_error());
while($row4 = mysql_fetch_array($result4))
{
echo $row4[diseasename];
}
}
}
?>
the line between ###doublestar### is the error line(49).

Just within index2.php :
$diseasetypeid = "$_POST[diseasetype]";
should be
$diseasetypeid = $_POST["diseasetype"];
and
if(isset($_REQUEST[submit]))
should really be this :
if($_SERVER['REQUEST_METHOD'] == 'POST')
this checks the REQUEST_METHOD rather than a form value. And
foreach ($_REQUEST[pets] as $key=>$values)
should be
foreach ($_REQUEST['pets'] as $key=>$values)
and I suggest you read about sql injection

This code:
foreach ($_REQUEST[pets] as $key=>$values)
Should be changed to:
foreach ($_REQUEST['pets'] as $key=>$values)
You should quote (put apostrophe or double-quotes, before and after your array key if it's string).
Then, you must have a form something to the following (i can't find it in your index.php file).
<input type="text" name="pets[]" />
<input type="text" name="pets[]" />
<input type="text" name="pets[]" />
<input type="text" name="pets[]" />
<input type="text" name="pets[]" />
The array key 'pets' is the name of your input. [] indicates array. In foreach ($something as ...), $something must be an array (or object).

Related

How to Update Value Based on the Displayed row in MySQL and PHP

I'm trying to update my data in MySQL database using the displayed data from it. But I haven't much luck in figuring out what I'm doing wrong. I don't know how can I pass my data from my $rows in order for me to use it in my UPDATE SQL statement.
I just want to complete all the data before I clicked the save button.
Please see attached image for the output..
And also, I got a warning message like this.
WHenever I Clicked the Save Data button..
Notice: Undefined variable: reason in C:\xampp\htdocs\BTR\get_forms.php on line 109
Notice: Array to string conversion in C:\xampp\htdocs\BTR\get_forms.php on line 109
Can someone enlighten me.. How to do this.? Thank you..
<?php
require 'config.php';
$form_type = $_POST['form_type'];
if ($form_type == 'MCCV-F2'){
$region = $_POST['region'];
$province = $_POST['province'];
$municipality = $_POST['municipality'];
$barangay = $_POST['barangay'];
$period = $_POST['period'];
$form_type = $_POST['form_type'];
echo "NON COMPLIANT IN EDUCATION<br>";
echo "<br><br>MUNICIPALITY: ".$municipality;
echo "<br><br>BARANGAY: ".$barangay;
echo "<br><br>PERIOD: ".$period;
?>
<form name="get_forms_f2" action="" method="post">
<br><br>
<center><table border = 1 style =2 width=1800>
<tr>
<td><center><b>Household ID </center></td>
<td><center><b>Member ID </center></td>
<td><center><b>Name</center></td>
<td><center><b>Sex</center></td>
<td><center><b>HH Status</center></td>
<td><center><b>Grade Level </center></td>
<td><center><b>School ID</center></td>
<td><center><b>Name Of Dominant School</center></td>
<td><center><b>CV Remarks</center></td>
<td><center><b>Reason</center></td>
<td><center><b>Other Reason</center></td>
<td><center><b>Intervention</center></td>
</tr>
<?php
$sql = "SELECT A.family_id, A.barangay, A.person_id, A.gender, A.family_status, A.current_grade_level,
A.school_facility_id, A.school_facility_name, A.municipality, CONCAT(B.last_name, ', ',B.first_name) as 'name',
B.person_id,B.cv_remarks, B.reason, B.other_reason, B.intervention, B.status FROM roster AS A RIGHT JOIN compliance AS B ON A.person_id = B.person_id
WHERE B.period='$period' AND B.form_type='$form_type' AND A.municipality='$municipality' AND A.barangay='$barangay'";
$query=$conn->prepare($sql);
$query->execute();
$result= $query->fetchALL(PDO::FETCH_ASSOC);
$count=(int)$query->rowCount();
foreach ($result as $row){
$person_id[] = $row['person_id'];
echo "<tr>";
echo "<td>".$row['family_id']."</td>";
echo "<td>".$row['person_id']."</td>";
echo "<td>".$row['name']."</td>";
echo "<td>".$row['gender']."</td>";
echo "<td>".$row['family_status']."</td>";
echo "<td>".$row['current_grade_level']."</td>";
echo "<td>".$row['school_facility_id']."</td>";
echo "<td>".$row['school_facility_name']."</td>";
echo "<td><input type='text' name='cv_remarks[]' value='".$row['cv_remarks']."'></td>";
echo "<td><select name='reason[]'>";
if (is_null($row['reason'])){
$sql2= "SELECT reason_code, reason_desc FROM reasons WHERE form_type ='2' ORDER BY reason_code ASC";
echo "<option value=''>SELECT REASON FOR Non-Compliance</option>";
foreach($conn->query($sql2) as $row2){
echo "<option value='".$row2['reason_desc']."'>".$row2['reason_code']." - ".$row2['reason_desc']."</option>";
}
}
if (!is_null($row['reason'])){
$sql2= "SELECT reason_code, reason_desc FROM reasons WHERE form_type ='2' ORDER BY reason_code ASC";
echo "<option value='".$row['reason']."'>".$row['reason']." (SELECTED)"."</option>";
foreach($conn->query($sql2) as $row2){
echo "<option value='".$row2['reason_desc']."'>".$row2['reason_code']." - ".$row2['reason_desc']."</option>";
}
}
echo "</select></td>";
echo "<td><input type='text' name='other_reason[]' value='".$row['other_reason']."'></td>";
echo "<td><input type='text' name='intervention[]' value='".$row['intervention']."'></td>";
echo "</tr>";
}
}
?>
</table></center><br><br>
<input type="submit" name="submit" value="Save Data">
<?php
$sql3 = "UPDATE compliance SET reason='{$reason}' WHERE person_id='{$person_id}' AND form_type='$form_type' AND period='$period'";
$query = $conn->prepare($sql3);
$query->execute();
?>
</form>
I have posted with the answer of how to update form with an array variable as a whole.
I have kept person,reason as array since you have added that alone as array().
<?php
require 'config.php';
if(isset($_POST['submit']))
{
for ($i=0; $i<count($_POST['person_id']); $i++)
{
$sql3 = "UPDATE compliance SET reason='".$_POST['reason'][$i]."' WHERE person_id='".$_POST['person_id'][$i]."' AND form_type='".$_POST['form_type']."' AND period='".$_POST['period']."'";
$query = $conn->prepare($sql3);
$query->execute();
}// end of For
}// end of IF
$form_type = $_POST['form_type'];
if ($form_type == 'MCCV-F2'){
$region = $_POST['region'];
$province = $_POST['province'];
$municipality = $_POST['municipality'];
$barangay = $_POST['barangay'];
$period = $_POST['period'];
echo "NON COMPLIANT IN EDUCATION<br>";
echo "<br><br>MUNICIPALITY: ".$municipality;
echo "<br><br>BARANGAY: ".$barangay;
echo "<br><br>PERIOD: ".$period;
?>
<form name="get_forms_f2" action="" method="post">
<br><br>
<center><table border = 1 style =2 width=1800>
<tr>
<td><center><b>Household ID </center></td>
<td><center><b>Member ID </center></td>
<td><center><b>Name</center></td>
<td><center><b>Sex</center></td>
<td><center><b>HH Status</center></td>
<td><center><b>Grade Level </center></td>
<td><center><b>School ID</center></td>
<td><center><b>Name Of Dominant School</center></td>
<td><center><b>CV Remarks</center></td>
<td><center><b>Reason</center></td>
<td><center><b>Other Reason</center></td>
<td><center><b>Intervention</center></td>
</tr>
<?php
$sql = "SELECT A.family_id, A.barangay, A.person_id, A.gender, A.family_status, A.current_grade_level,
A.school_facility_id, A.school_facility_name, A.municipality, CONCAT(B.last_name, ', ',B.first_name) as 'name',
B.person_id,B.cv_remarks, B.reason, B.other_reason, B.intervention, B.status FROM roster AS A RIGHT JOIN compliance AS B ON A.person_id = B.person_id
WHERE B.period='$period' AND B.form_type='$form_type' AND A.municipality='$municipality' AND A.barangay='$barangay'";
$query=$conn->prepare($sql);
$query->execute();
$result= $query->fetchALL(PDO::FETCH_ASSOC);
$count=(int)$query->rowCount();
foreach ($result as $row){
?>
<input type="hidden" name="person_id[]" value="<?php echo $row['person_id'];?>">
<input type="hidden" name="form_type" value="<?php echo $form_type; ?>">
<input type="hidden" name="period" value="<?php echo $period; ?>">
<?php
echo "<tr>";
echo "<td>".$row['family_id']."</td>";
echo "<td>".$row['person_id']."</td>";
echo "<td>".$row['name']."</td>";
echo "<td>".$row['gender']."</td>";
echo "<td>".$row['family_status']."</td>";
echo "<td>".$row['current_grade_level']."</td>";
echo "<td>".$row['school_facility_id']."</td>";
echo "<td>".$row['school_facility_name']."</td>";
echo "<td><input type='text' name='cv_remarks[]' value='".$row['cv_remarks']."'></td>";
echo "<td><select name='reason[]'>";
if (is_null($row['reason'])){
$sql2= "SELECT reason_code, reason_desc FROM reasons WHERE form_type ='2' ORDER BY reason_code ASC";
echo "<option value=''>SELECT REASON FOR Non-Compliance</option>";
foreach($conn->query($sql2) as $row2){
echo "<option value='".$row2['reason_desc']."'>".$row2['reason_code']." - ".$row2['reason_desc']."</option>";
}
}
if (!is_null($row['reason'])){
$sql2= "SELECT reason_code, reason_desc FROM reasons WHERE form_type ='2' ORDER BY reason_code ASC";
echo "<option value='".$row['reason']."'>".$row['reason']." (SELECTED)"."</option>";
foreach($conn->query($sql2) as $row2){
echo "<option value='".$row2['reason_desc']."'>".$row2['reason_code']." - ".$row2['reason_desc']."</option>";
}
}
echo "</select></td>";
echo "<td><input type='text' name='other_reason' value='".$row['other_reason']."'></td>";
echo "<td><input type='text' name='intervention' value='".$row['intervention']."'></td>";
echo "</tr>";
}
}// end of MCCV-F2 form
?>
</table>
</center>
<br><br>
<input type="submit" name="submit" value="Save Data">
</form>
Hope so this would solve your problem br. have a try and let me know if any hindrance you find.

How do I run multiple SQL Queries using "if(isset($_POST['Submit'])){"

Trying to make a CRUD, everything works except my Update function. I feel like the problem is in the second sql query. When I click on submit it just refreshes and the change is gone. Can anyone show me how to find what I need to change/show me what to change?
<head>
<title>Update</title>
</head>
<body>
</form>
<?php
require_once('dbconnect.php');
$id = $_GET['id'];
$sql = "SELECT * FROM dealers where ID=$id";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo '<form action="" method="post">';
echo "Company: <input type=\"text\" name=\"CName\" value=\"".$row['CName']."\"></input>";
echo "<br>";
echo "Contact: <input type=\"text\" name=\"Contact\" value=\"".$row['Contact']."\"></input>";
echo "<br>";
echo "City: <input type=\"text\" name=\"City\" value=\"".$row['City']."\"></input>";
echo "<br>";
echo "<input type=\"Submit\" = \"Submit\" type = \"Submit\" id = \"Submit\" value = \"Submit\">";
echo "</form>";
}
echo "</table>";
} else {
echo "0 results";
}
if(isset($_POST['Submit'])){
$sql = "UPDATE dealers SET CName='$CName', Contact='$Contact', City='$City' where ID=$id";
$result = $conn->query($sql);
}
$conn->close();
?>
Instead of building a form inside PHP, just break with ending PHP tag inside your while loop and write your HTML in a clean way then start PHP again. So you don't make any mistake.
Also you've to submit your $id from your form too.
Try this
<?php
require_once('dbconnect.php');
$id = $_GET['id'];
$sql = "SELECT * FROM dealers where ID=$id";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
?>
<form action="" method="post">
<input type="hidden" name="id" value="<?= $id ?>" />
Company: <input type="text" name="CName" value="<?= $row['CName'] ?>" />
<br>
Contact: <input type="text" name="Contact" value="<?= $row['Contact'] ?>" />
<br>
City: <input type="text" name="City" value="<?= $row['City'] ?>" />
<br>
<input type="Submit" name="Submit" id="Submit" value="Submit" />
</form>
<?php
} // end while loop
echo "</table>";
}
else {
echo "0 results";
}
Note: You are passing undefined variables into your update query. As you are submitting your form you must have to define those variables before you use them.
if (isset($_POST['Submit'])) {
$CName = $_POST['CName'];
$Contact = $_POST['Contact'];
$City = $_POST['City'];
$id = $_POST['id'];
$sql = "UPDATE dealers SET CName='$CName', Contact='$Contact', City='$City' where ID=$id";
$result = $conn->query($sql);
}
$conn->close();
that loop? ID primary key or not?
maybe u need create more key in table dealer like as_id
<input type="hidden" name="idform" value="$as_id">
in statment
if($_POST){
$idf = $_POST['idform'];
if(!empty($idf)){
$sql = "UPDATE dealers SET CName='$CName', Contact='$Contact', City='$City' where as_id=$idf";
$result = $conn->query($sql);
}
$conn->close();
}

Display saved checkbox value

I have some checkbox options that I save in the DB. I was able to view and also select multiple options and save them in the DB. The issue is that I want to display the saved information but I don't know how to do that.
<form action="save_comp.php" method="post">
<?php
//Display
include ('mysql_connect.php');
$sql = mysql_query("SELECT * FROM competency ");
//$row = mysql_fetch_array($sql);
while($row = mysql_fetch_array($sql))
{
echo"<input type='checkbox' name='comp[]' value= ".$row['id']." /> ".$row['competency']." <br />";
}
?>
<input name="submit" type="submit" value="submit" />
</form>
Save into DB
<?php
session_start();
$id = $_SESSION['user_id'];
//$id = 3;
include ('mysql_connect.php');
$insStr = '';
foreach($_POST['comp'] as $val){ $insStr .=$val.","; }
mysql_query("INSERT INTO competency_result (user_id,result) VALUES ( '$id', '$insStr' )") or die(mysql_error());
echo'<script>alert("Inserted Successfully")</script>';
?>
All I want to do now is to display the saved information in a table format. I tried doing this but it only showed me the saved ID
<?php
$res= mysql_query("SELECT * FROM competency_result WHERE user_id = '$user'")or die(mysql_error());
while($row = mysql_fetch_array($res))
{
echo"<tr>";
echo"<td> $row[result]</td>";
?>
<?php
echo"</tr>";
}
?>
<form action="save_comp.php" method="post">
<?php
//Display
include ('mysql_connect.php');
$sql = mysql_query("SELECT * FROM competency ");
//$row = mysql_fetch_array($sql);
while($row = mysql_fetch_array($sql))
{
echo"<input type='checkbox' name='comp[". $row['id']. "]' value='". $row['competency'] ."' /> ".$row['competency']." <br />";
}
?>
<input name="submit" type="submit" value="submit" />
</form>
If you want to checkboxes check then you can try with below code:
<?php
$sql = mysql_query("SELECT name FROM competency ");
//$row = mysql_fetch_array($sql);
while($row = mysql_fetch_array($sql))
{
$focus=explode(",",$row['name']);
?>
<input type="checkbox" name="focus[]" value="Art" <?php if(in_array("Comp",$focus)) { ?> checked="checked" <?php } ?> >
<input type="checkbox" name="focus[]" value="Mathematics" <?php if(in_array("Mathematics",$focus)) { ?> checked="checked" <?php } ?> >
<input type="checkbox" name="focus[]" value="Dance" <?php if(in_array("Dance",$focus)) { ?> checked="checked" <?php } ?> >
<?php
}
?>

Update echoed data using WHILE loop. Only updates one record

I can't seem to be able to update any records except the first one.
I am not sure how to modify any of the displayed records.
<?php
if(isset($_POST["action"]) == "update")
{
$id = $_POST['m_id'][0];
$type = $_POST['type'][0];
// if I echo $id & $type, it only gives me the first record.**
mysql_query("
UPDATE membership_type
SET mt_type ='$type'
WHERE mt_id = '$id'"
);
}
?>
ALl of this is within the same php page.
<form name=form action='' method='post'>
<?php
$result=mysql_query("SELECT * FROM membership_type;");
while($rows=mysql_fetch_array($result))
{ ?>
<input size=35 class=textField type=text name='type[]' value='<?php echo $rows['mt_type']; ?>'>
<input type=hidden name='m_id[]' value="<?php echo $rows['mt_id']; ?>">
<input type=submit value="Update">
<?php
}
?>
How do I edit any of the displayed records by simply clicking Update button???
First: You should NEVER use the mysql_* functions as they are deprecated.
Second: Try this code:
<?php
// Get a connection to the database
$mysqli = new mysqli('host', 'user', 'password', 'database');
// Check if there's POST request in this file
if($_POST){
foreach($_POST['m_id'] as $id => $type){
$query = "UPDATE membership_type
SET mt_type = '".$type."'
WHERE mt_id = '".$id."'";
// Try to exec the query
$mysqli->query($query) or die($mysqli->error);
}
}else{
// Get all membership_type records and then iterate
$result = $mysqli->query("SELECT * FROM membership_type") or die($mysqli->error); ?>
<form name='form' action='<?php echo $_SERVER['PHP_SELF'] ?>' method='post'>
<?php while($row = $result->fetch_object()){ ?>
<input size='35'
class='textField'
type='text'
name='m_id[<?php echo $row->mt_id ?>]'
value='<?php echo $row->mt_type; ?>'>
<input type='submit' value="Update">
<?php } ?>
</form>
<?php } ?>
Third: In order to add more security (this code is vulnerable), try mysqli_prepare
Only the first record is updated on every form submission because you have set $id = $_POST['m_id'][0], which contains the value of the first type[] textbox. To update all the other records as well, loop through $_POST['m_id'].
Replace it. Hope this works.
<?php
if(isset($_POST["action"]) == "update")
{
$id = $_POST['m_id'];
$type = $_POST['type'];
$i = 0;
foreach($id as $mid) {
mysql_query("UPDATE membership_type
SET mt_type='".mysql_real_escape_string($type[$i])."'
WHERE mt_id = '".intval($mid)."'") OR mysql_error();
$i++;
}
}
?>
Try this :
if(isset($_POST["action"]) == "update")
{
$id = $_POST['m_id'];
$type = $_POST['type'];
$loopcount = count($id);
for($i=0; $i<$loopcount; $i++)
{
mysql_query("
UPDATE membership_type
SET mt_type ='$type[$i]'
WHERE mt_id = '$id[$i]'"
);
}
}
You HTML was malformed and you were passing as an array but then only using the first element. Consider:
<form name="form" action="" method="post">
<?php
$result = mysql_query("SELECT * FROM membership_type;");
while($row = mysql_fetch_array($result))
echo sprintf('<input size="35" class="textField" type="text" name="m_ids[%s]" value="%s" />', $row['mt_id'], $row['mt_type']);
?>
<input type="submit" value="Update">
</form>
Then the server script:
<?php
if(isset($_POST["action"]) && $_POST["action"] == "Update"){
foreach($_POST['m_ids'] as $mt_id => $mt_type)
mysql_query(sprintf("UPDATE membership_type SET mt_type ='%s' WHERE mt_id = %s LIMIT 1", addslashes($mt_type), (int) $mt_id));
}
There are other things you could be doing here, eg. prepared statements, but this should work.

checked values for the SELECT: WHERE clause

I am trying to create a compare option between selected cars.
<?php
if(isset($_POST['compares'])) {
$id_nums = array($_POST['cBox']);
//$id_nums = array(1,6,12,18,24);
$id_nums1 = implode(", ", $id_nums);
$query = "SELECT * FROM wp_cars WHERE id in ($id_nums1)";
$cCars = mysql_query($query) or mysql_error();
while($car = mysql_fetch_array($cCars)) {
echo $car['cartitle']."<br/>";
echo $car['saleprice']."<br/>";
}
} else {
$query1 = "SELECT * FROM wp_cars";
$allcars = mysql_query($query1) or die(mysql_error()); `
while($car1 = mysql_fetch_array($allcars)) {
echo "<input type='checkbox' value=".$car1['id']." name='cBox[]' />";
echo $car1['cartitle']."<br/>";
echo $car1['saleprice']."<br/>";
}
}
?>
How to pass the checkbox name(cBox[]) array based on checkboxes selection.
<form action="compares.php" method="post">
<button name="compares">Select Cars to Compare</button>
</form>
$id_nums = array($_POST['cBox']);
$_POST['cBox'] is already an array, you are making a 2d array. Doing
$id_nums1 = implode(", ", $_POST['cBox']);
would do what you want. Although it is wide open to SQL injection.
From your HTML part, send inputs this way:
<form action="compares.php" method="post">
<?php foreach (mysql_fetch_array($allcars) as $car: ?>
<input type="checkbox" value="<?php echo $car['id']; ?>" name="cBox[]" />
<?php echo $car['cartitle']; ?><br />
<?php echo $car['saleprice']; ?><br />
<?php endforeach; ?>
</form>
And in your PHP part, receive inputs this way:
$id_nums = implode(",", $_POST['cBox']);

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