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update query is not working , giving Mariadb server version error

shruti A
shruti A asked
Last Modified: 2017-04-06
include ('db.php'); 
include ('header.php'); 

<div class="container">
<div class="">
 <div class="container-con">
<!-- end banner & menunav -->

<div class="container">
<div class="">
<div class="span12">
<div class="row-fluid">
<div class="span3"></div>
<div class="span6">

<div class="hero-unit-3">

$sql="select * from combo1 where id='$id'";
  $query=mysqli_query($link,$sql)or die(mysqli_error($link));
 <form class="form-horizontal" method="post"  enctype="multipart/form-data">
								<div class="control-group">
                                    <label class="control-label" for="inputColumn">COLUMN:</label>
                                    <div class="controls">
                                        <input type="text" name="column" required value=<?php echo $row['column']; ?>>
								<div class="control-group">
                                    <label class="control-label" for="inputType">TYPE:</label>
                                    <div class="controls">
                                        <input type="text" name="type" required value=<?php echo $row['type']; ?>>
                                <div class="control-group">
                                    <label class="control-label" for="inputValue">VALUE:</label>
                                    <div class="controls">
                                        <input type="text" name="value" required value=<?php echo $row['value']; ?>>
								 <div class="control-group">
                                    <div class="controls">

                                        <button type="submit" name="update" class="btn btn-success" style="margin-right: 65px;" value="update">Save</button>
										<a href="excombo.php" class="btn">Back</a>


$result = mysqli_query($link,"SELECT * FROM combo1 WHERE id = '$id'");
$test = mysqli_fetch_array($result);
if (!$result) 
		die("Error: Data not found..");
									$column= $test['column'];
									$type= $test['type'];
									$value= $test['value'];
if (isset($_POST['update'])) {

									// $column=(isset( $_POST['column']));
									$column = isset( $_POST['column']) ? $_POST['column']:" ";
									//$type= (isset($_POST['type']));
									$type = isset( $_POST['type']) ? $_POST['type']:" ";
									//$value= (isset($_POST['value']));
									$value = isset( $_POST['value']) ? $_POST['value']:" ";
									echo $id;

									echo $column;
									echo $type;
									echo $value;
mysqli_query($link,"UPDATE combo1 SET column = '$column' type = '$type'  value ='$value' WHERE id ='$id'") 
or die(mysqli_error($link)); 




Open in new window

it ia giving error like this
25col1INTAddressYou have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'column = 'col1' type = 'INT' value ='Address' WHERE id ='25'' at line 1
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