I am new to StackOverflow, and relatively new to PHP. I have a mixture of a mySQL table and form input, where a user will enter numbers into a field (options) next to each corresponding record (companyid), then submit and these should be inserted into the mySQL table. If no "options" are entered, the loop should skip and move on.
When I submit, nothing is actually inserted into the database. I need a little help because I have looked at several examples and I thought I had everything correct.
Thank you so much!!
Below is the table with form inputs:
<?php
$result = mysqli_query($connection,"SELECT * FROM company_main");
echo "<form action='userinvestdynamic.php' method='post'>";
while($row = mysqli_fetch_array($result))
{
echo "<table border=1>";
echo "<tr>";
echo "<td>Company ID</td>";
echo "<td>" .$row['company_id'] . "</td>";
echo "<input type=hidden name=companyid[] value=" . $row['company_id'] . " />";
echo "<td><input type=number name=options[]></td>";
echo "</tr>";
echo "</table>";
}
echo "<input name='userinvestoptionsdynamic' type='submit' value='Invest!'>";
echo "</form>";
?>
And here is the code once the form is submitted.
<?php
require ('db_login.php');
session_start();
if (isset($_POST['userinvestoptionsdynamic'])) {
$userid = $_SESSION['login_user'];
$companyid = $_POST['companyid'];
$options = $_POST['options'];
// loop through array
$number = count($_POST['companyid']);
for ($i=0; $i<$number; $i++)
{
//store single company and options in local variables
$companyid = $companyid[$i];
$options = $options[$i];
//run insert for any items that don't have blank options
if($options[$i] <>''){
$query = "INSERT INTO user_company_invested(user_id, company_id, user_company_options_invested) VALUES($userid,$companyid,$options)";
mysqli_query($connection, $query);
}
}
header('Location: userpage.php');
} else {
echo 'error';
}
?>
Explanations are inside the comment /* */
<?php
$result = mysqli_query($connection,"SELECT * FROM company_main");
$counter=0; /* DECLARE THIS BEFORE YOUR WHILE LOOP */
echo "<form action='userinvestdynamic.php' method='post'>";
while($row = mysqli_fetch_array($result))
{
echo "<table border=1>";
echo "<tr>";
echo "<td>Company ID</td>";
echo "<td>" .$row['company_id'] . "</td>";
echo "<input type=hidden name=companyid[$counter] value=" . $row['company_id'] . " />"; /* DECLARE THE COUNTER ON COMPANY ID ARRAY */
echo "<td><input type=number name=options[$counter]></td>"; /* DECLARE THE COUNTER ON OPTIONS ARRAY */
echo "</tr>";
echo "</table>";
$counter=$counter+1; /* ADD 1 EVERY LOOP ON COUNTER */
}
echo "<input type='hidden' name='hiddencounter' value='$counter'>"; /* COUNT ALL THE LOOP AND SUBMIT IT TO THE NEXT FORM */
echo "<input name='userinvestoptionsdynamic' type='submit' value='Invest!'>";
echo "</form>";
?>
userinvestdynamic.php:
<?php
require ('db_login.php');
session_start();
if (isset($_POST['userinvestoptionsdynamic'])) {
$userid = $_SESSION['login_user']; /* I ASSUME YOU HAVE LOGIN_USER SESSION DECLARED IN THE PREVIOUS FILE? */
$companyid = $_POST['companyid'];
$options = $_POST['options'];
$counter = $_POST['hiddencounter']; /* THIS WILL SET AS YOUR COUNTER */
for ($i=0; $i<=$counter; $i++)
{
if(empty($options[$i])){ /* IF NO INPUT ON OPTIONS */
/* DO NOTHING */
}
else {
/* store single company and options in local variables */
$query = "INSERT INTO user_company_invested(user_id, company_id, user_company_options_invested)
VALUES($userid,$companyid[$i],$options[$i])";
mysqli_query($connection, $query);
} /* END OF ELSE IF NOT EMPTY COMPANY ID */
} /* END OF FOR LOOP */
header('Location: userpage.php');
} /* END OF ISSET USERINVESTOPTIONSDYNAMIC */
else {
echo 'Error!';
}
?>
Related
I have been trying to work this out for a while.
I have created a form with a dropdown box that gets results from a database. from this i then $_POST that from to another page. From that second page i wish to get the ID number and then get the records and display them on screen.
I will then put them in a table to organise the results better.
can anyone help me in achieving this.
Here is the code for the form (which works and sends the $PlantID)
$sql = "SELECT DISTINCT * FROM PLANTS";
$result = mysqli_query($mysqli,$sql)or die(mysqli_error());
//********************* Botannical name drop down box
echo "<form name='selection' id='selection' action='profile.php' method='post'>";
echo "<select name='flower'>";
while($row = mysqli_fetch_array($result)) {
$plantid = $row['FlowerID'];
$plantname = $row['Botannical_Name'];
$plantcommon = $row['Common_Name'];
/* $plantheight = $row['Height'];
$plantav = $row['AV'];
$plantcolours = $row['Colours'];
$plantflowering = $row['Flower_Time'];
$plantspecial = $row['Special_Conditions'];
$plantfrost = $row['Frost_Hardy'];
$plantaspect = $row['Aspect'];
$plantspeed = $row['Growth_Speed'];*/
echo "<option value=".$plantid.">".$plantname." -> AKA -> ".$plantcommon."</option>";
}
echo "</select>";
echo "<br />";
//********************* End of form
echo "<input type='submit' name='submit' value='Submit'/>";
echo "</form>";
I have created this page to get the ID and display that ID on screen. AS you can tell i have probably doubled up on ways to try work this out.
$sql = "SELECT * FROM PLANTS";
$result = mysqli_query($mysqli,$sql)or die(mysqli_error());
if(isset($_POST['submit'])){
$selected_val = $_POST['Botannical_Name']; // Storing Selected Value In Variable
echo "You have selected :" .$selected_val; // Displaying Selected Value
}
echo "<br />";
echo "well:".$_POST["Botannical_Name"]."<br/>";
echo "now:".$plantquery."<br />";
echo $_POST;
echo "<table>";
foreach ($_POST as $key => $value) {
echo "<tr>";
echo "<td>";
echo $key;
echo "</td>";
echo "<td>";
echo $value;
echo "</td>";
echo "</tr>";
}
echo "</table>";
Any help would be greatly appreciated.
you should use following to get the selected value,
$selected_val = $_POST['flower'];
if(isset($_POST['submit'])){
$selected_val = $_POST['flower']; // Storing Selected Value In Variable
echo "You have selected :" .$selected_val; // Displaying Selected Value
$sql = "SELECT * FROM PLANTS WHERE FlowerID='.$selected_val.'";
$result = mysqli_query($mysqli,$sql)or die(mysqli_error());
while ($row=mysqli_fetch_assoc($result))
{
echo $row['Botannical_Name'];
}
}
echo "<br />";
print_r($_POST);
if(!empty(_POST)) {
echo "<table>";
foreach ($_POST as $key => $value) {
echo "<tr>";
echo "<td>";
echo $key;
echo "</td>";
echo "<td>";
echo $value;
echo "</td>";
echo "</tr>";
}
echo "</table>";
}
I am trying to get the column names and data from a table using mysqli and php. Any help is appreciated.
I know how to do this in java by more or less this code:
String [] header= new String[numberOfColumns-1];
//Code to get column names in header array
String table = "<table><tr>"
for(int i=0;i<numberOfColumns-1;i++){
table= table + "<td>" + header[i] + </td>;
}
table = table + </tr>;
while(result.next()!=null){
table = table + <tr>
for (int j = 0 ; j < numberOfColumns-1 ; j++){
table = table + <td> + result.getString[j] + </td>
}
table = table + </tr>
}
But i have no idea how to do it in php. Each table is different but what i have to far for the one table:
include("config.php");
session_start();
$sql = ($_POST['q']);
$result = mysqli_query($db,$sql);
echo "<tr>";
echo "<td>First Name</td>";
echo "<td>Last Name</td>";
echo "<td>Date of Birth</td>";
echo "<td>Contact Details</td>";
echo "</tr>";
while($rowitem = mysqli_fetch_array($result,MYSQLI_ASSOC)){
echo "<tr>";
echo "<td>" . $rowitem['fname'] . "</td>";
echo "<td>" . $rowitem['lname'] . "</td>";
echo "<td>" . $rowitem['dob'] . "</td>";
echo "<td>" . $rowitem['contact'] . "</td>";*/
echo "</tr>";
}
echo "</table>"; //end table tag
EDIT: Is the implementation of the fetch_fields correct?
include("config.php");
session_start();
$sql = ($_POST['q']);
$result = mysqli_query($db,$sql);
$finfo = mysqli_fetch_fields($result);
echo "<tr>";
foreach ($finfo as $val) {
echo "<td>" . $val->name . "</td>"
}
echo "</tr>";
while($rowitem = mysqli_fetch_array($result,MYSQLI_ASSOC)){
echo "<tr>";
for($x =0; $x < count($finfo);$x++){
echo "<td>" . $rowitem[$x] . "/td";
}
echo "</tr>";
}
echo "</table>";
EDIT 2: Database Connection.php
<<?php
include("config.php");
session_start();
$sql = ($_GET['q']);
$result = mysqli_query($db,$sql);
echo "<table border='1' class='table_info'>";
$finfo = mysqli_fetch_fields($result);
echo "<tr>";
foreach ($finfo as $val) {
echo "<td>" . $val->name . "</td>";
}
echo "</tr>";
while($rowitem = mysqli_fetch_array($result,MYSQLI_ASSOC)){
echo "<tr>";
foreach ($row as $value) { (line 18)
echo "<td>" . $value . "</td>";
}
echo "</tr>";
}
echo "</table>"; //end table tag
?>
Error:
Notice: Undefined variable: row in databaseConnection.php on line 18
Warning: Invalid argument supplied for foreach() in databaseConnection.php on line 18
You can call mysqli_fetch_fields on the $result object, and retrieve the name property for each object in the returned array. See documentation for examples.
There are a couple ways to do this:
Use array_keys() to grab the names of the keys from an associative array:
$rowitem = mysqli_fetch_array($result, MYSQLI_ASSOC);
$header = array_keys($rowitem);
Use mysqli_fetch_fields() to grab the field objects, and parse out the names into an array using array_map():
// Helper function to use in array_map
// (For PHP < 5.3. If >= 5.3.0, use the second version)
function getFieldName($field) { return $field->name; }
// For PHP < 5.3:
$header = array_map("getFieldName", mysqli_fetch_fields($result));
// For PHP >= 5.3.0:
$header = array_map(function($o) { return $o->name; }, mysqli_fetch_fields($result));
As mentioned you can use $result->fetch_fields() to get the field information from the result and this includes the name of each field.
For what you're trying to acheive this code should work for pretty much any table/query:
<?php
/* Connect to mysql */
$mysqli = new mysqli("localhost", "username", "password", "database");
/* Check connection */
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
/* Your query */
$query = "SELECT * from table";
if ($result = $mysqli->query($query)) {
/* Output HTML table */
echo "<table border='1'>";
/* Get field information for all returned columns */
$finfo = $result->fetch_fields();
/* Output each column name in first table row */
echo "<tr>";
foreach ($finfo as $val) {
echo "<td>".$val->name."</td>";
}
echo "</tr>";
/* Fetch the result and output each row into a table row */
while($row = mysqli_fetch_array($result, MYSQLI_NUM)) {
echo "<tr>";
foreach ($row as $value) {
echo "<td>" . $value . "</td>";
}
echo "</tr>";
}
$result->free();
echo "</table>";
}
$mysqli->close();
(Obviously you should never query a database from a $_POST variable without some serious sanitisation of the input, but I'll assume that was just for your convenience.)
You can Echo in a table..
$result = mysqli_query($db,$sql)
if (!$result) {
die("Query to show fields from table failed");
}
$fields_num = mysqli_num_fields($result);
echo "<table border='1'><tr>";
// printing table headers
for($i=0; $i<$fields_num; $i++)
{
$field = mysqli_fetch_field($result);
echo "<td>{$field->name}</td>";
}
echo "</tr>\n";
// printing table rows
while($row = mysqli_fetch_row($result))
{
echo "<tr>";
// $row is array... foreach( .. ) puts every element
// of $row to $cell variable
foreach($row as $cell)
echo "<td>$cell</td>";
echo "</tr>\n";
}
mysqli_free_result($result);`
Problem is while loop always showing last inserted row result. I've tried this below code to follow/unfollow button option. For example I'm a user id=1. I have already followed user id=4. Now, I want to follow user id=5. When i click follow button(id=5) it turns into Unfollow properly. But, I have already followed user id=4. That turns into Follow. This is my problem.
Then I tried echo $following;. it Prints 5555. That means last inserted data. But I want 45. I'm sure I've made a mistake in my while loop. But I don't know what I should change?
<?php
try
{
$stmt = $conn->prepare("SELECT * FROM users ORDER BY Autoid");
$stmt->errorInfo();
$stmt->execute();
$sth = $conn->prepare("SELECT * FROM followers ORDER BY Autoid");
$sth->errorInfo();
$sth->execute();
while($follow_row = $sth->fetch(PDO::FETCH_ASSOC))
{
$following = $follow_row['Following'];
$follower = $follow_row['Follower'];
}
while ($row = $stmt->fetch(PDO::FETCH_ASSOC))
{
echo "<tr>";
echo "<td>". $row['Autoid'] ."</td>";
echo "<td>". $row['Name'] ."</td>";
// echo $row['Following'];
if($_SESSION['sesuname'] == $row['Username'])
{
echo "<td class='itsyou' >Its You ". $_SESSION['sesuname'] ."</td>";
}
else
{
if(($follower == $_SESSION['sesid']) AND ($following != $row['Autoid']))
{
//echo "<td>true</td>";
echo $following;
echo "<td>";
echo "<form id='jsform' method='post' action='subscribe.php'>";
echo "<input type='hidden' name='id' value=" . $row['Autoid'] . " />";
echo "<button class='follow' >Follow</button>";
echo "</form>";
echo "</td>";
}
else
{
//echo "<td>false</td>";
echo "<td>";
echo "<form id='ufform' method='post' action='unsubscribe.php'>";
echo "<input type='hidden' name='uid' value=" . $row['Autoid'] . " />";
echo "<button class='follow' >UnFollow</button>";
echo "</form>";
echo "</td>";
}
}
echo "</tr>";
}
} catch (PDOException $e) {
'Database Error : ' .$e->getMessage();
}
?>
This code:
while($follow_row = $sth->fetch(PDO::FETCH_ASSOC))
{
$following = $follow_row['Following'];
$follower = $follow_row['Follower'];
}
Simply OVERWRITES $following and $follower every time you fetch a row, leaving you with the LAST row fetched in the variables. Perhaps you want something more like
$following[] = $follow_row['Following'];
$follower[] = $follow_row['Follower'];
^^--- append new row value to an array.
I'm having problems to find how to create an hyperlink in a table column result and then, on click, open another page with all fields (textboxes) filled. Imagine when a click an ID, i do a select * from table where column_id = ID... Is there a way to do it?
Thanks.
Best regards
I'm not completely sure what you are asking, but this may help you a bit.
First make a Javascript.
<script type="text/JavaScript">
function selectID() {
var ID = document.getElementById("ID").value;
document.location.href ="yoursite.php?ID="+ID;
}
</script>
Then connect to your database to query the table for a link ID (or more) for example by changing the variable $value.
<?php
//Connect to database
mysql_connect("host", "user", "pass");
mysql_select_db("db_name");
$value = 'something';
$ID = $_GET['ID'];
if (!$ID) {
$ID = 0;
}
if ($ID == 0) {
$query = "SELECT * FROM table WHERE `column_1` = '$value'";
$result = mysql_query($query);
echo "<table>";
while($myrow = mysql_fetch_array($result)) {
echo "<tr>";
echo "<td>";
echo "ID";
echo "</td>";
echo "</tr>";
}
echo "</table>";
}
elseif ($ID > 0) {
$query2 = "SELECT * FROM table WHERE `column_id` = '$ID'";
$result2 = mysql_query($query2);
while($myrow2 = mysql_fetch_array($result2)) {
$value1 = $myrow2['column_1'];
$value2 = $myrow2['column_2'];
}
echo "<form type=\"GET\" action=\"$PHP_SELF\">";
echo "<input type=\"text\" id=\"ID\" name=\"ID\" value=\"$ID\"><br>";
echo "<input type=\"text\" id=\"value1\" name=\"value1\" value=\"$value1\"><br>";
echo "<input type=\"text\" id=\"value2\" name=\"value2\" value=\"$value2\"><br>";
echo "<input type=\"hidden\" id=\"search\" name=\"search\" value=\"searching\">";
echo "<input type=\"submit\" id=\"submitbutton\" name=\"submitbutton\" value=\" Search \">";
echo "</form>";
}
?>
I've been trying think of a way to do this. I want it to where users can check off items, hit submit and it goes to the code on the next page and deletes all of the checked items from the database. Problem one is that in the post its only sending over the last checked item. Here is how I have it set up right now.
echo "<form name='fm1' METHOD ='POST' ACTION ='displaydelete.php' > ";
//Draws up the table headers
echo "";
echo "";
echo "Fund Number ";
echo "Hours ";
echo "Percentage";
echo "Delete";
echo "";
//While there are query results data is pushed into table cells
while ($row = mysql_fetch_array($queryResult2))
{
$hours = $row['hours'];
$percentage = $hours / 160 * 100;
echo "<tr>";
echo "<td>";
echo $row['funnumber'];
echo "</td>";
echo "<td>";
echo $hours;
echo "</td>";
echo "<td>";
echo $percentage ."%";
echo "</td>";
echo "<td>";
echo "<input type='checkbox' name='id' value='$row[id]'/>";
echo "</td>";
echo "</tr>";
}
//End of tabel
echo "</table>";
echo" ";
echo "";
What I would like to do is push all of the items into a variable and maybe delete them that way. I'm not really sure how you would handle multiple deletes. I'm doing my delete like this for something else if this helps any.
$query = "DELETE FROM users
WHERE ninenumber = '$ninenumber'";
$result = mysql_query($query)
or die("Query Failed: " .mysql_error());
mysql_close($conn);
In your form:
<input type='checkbox' name='id[]' value='$row[id]'/>
Then, in the file you post to:
if(is_array($_POST['id'])){
foreach($_POST['id'] as $id){
...do something to $id;
}
}
Instead of this:
echo "<input type='checkbox' name='id' value='$row[id]'/>";
You need this:
echo "<input type='checkbox' name='id[]' value='$row[id]'/>";
Note the difference. I added [] after the input name. This tells the client and server that there are multiple inputs with that name. $_POST['id'] will be an array you can loop through on the next page.
foreach ($_POST['id'] as $checkbox) {
// DELETE FROM users WHERE ninenumber = $checkbox
}
isset, is_array, and mysql_real_escape_string omitted for brevity.
In the form-generating code, make the name in the html have" []" after it:
...
echo "<input type='checkbox' name='id[]' value='$row[id]'/>";
...
Then, in the form-reading code, your post'ed id will be an array.
$id_array = isset($_POST['id']) && is_array($_POST['id']) ? $_POST['id'] : array();
foreach( $id_array as $id ) {
$query = "DELETE FROM users WHERE ninenumber = '" . mysql_real_escape_string($id) . "'";
// execute the delete query
}
Putting [] after the name of a control will turn it into an array in the superglobal that you can then iterate over to get all the values from.
You need to have the same name for all of your checkboxes, then all values are passed as array POST variable.
<input type='checkbox' name='id[]' value='$row[id]'/>
Change
name=id
to
name=id[]
this will then give you an array.