sql语句，其中part等于null？

Question

In where part of sql statement id=".$_REQUEST['id'] ." ";value is feed from combobox that show the respective record.On page load the combobox has no value selected so the sql statement shows sql error on where part(id=".$_REQUEST['id'] .")?How to remove this error .

                <div class="col-lg-6" style="display:none"  id="d" > 




                                <select name="id" id="id" class="span2" style="width:150px;"  onchange="this.form.submit();">
                            <?php 
                    $servername = "localhost";
                    $username = "root";
                    $password = "";
                    $dbname = "valet";

                    // Create connection
                    $conn = mysqli_connect($servername, $username, $password, $dbname);
                    // Check connection
                    if (!$conn) {
                        die("Connection failed: " . mysqli_connect_error());
                    }
                    $age = array();

                    $sql = "SELECT * FROM `tbl_companies`  ";
                    $result = mysqli_query($conn, $sql);

                       // echo "User name=" . $row["name"]. "<br>";



                 ?>
                <option value="">-select company-</option>
                    <?php                   if (mysqli_num_rows($result) > 0) {
                        // output data of each row
                        while($row = mysqli_fetch_assoc($result)) {
                             $age[] = $row;
                             }
                               ?>


                <?php foreach($age as $key=>$row) {

                echo "<option value='" . $row['id'] . "' " . (($_REQUEST['id'] == $row['id']) ? "selected": "") . ">" . $row["id"] . "</option>";

                    ?>
                <?php } 
                   }
 else {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    header('Location: webservices.php');
}

mysqli_close($conn);
                ?>
            </select>           

            <form action="" id="abc"  method="post"   >

                    <br><br>
                <br>

<?php 
                    $servername = "localhost";
                    $username = "root";
                    $password = "";
                    $dbname = "valet";

                    // Create connection
                    $conn = mysqli_connect($servername, $username, $password, $dbname);
                    // Check connection
                    if (!$conn) {
                        die("Connection failed: " . mysqli_connect_error());
                    }
                    $age = array();

if(strlen($_REQUEST['id']) > 0)
{
                     $sql = "SELECT  company_name ,acess_code ,register_on  FROM `tbl_companies` where id=".$_REQUEST['id'] ." ";

                     $result = mysqli_query($conn, $sql);

                       // echo "User name=" . $row["name"]. "<br>";



if (mysqli_num_rows($result) > 0) {
                        // output data of each row
                        while($row = mysqli_fetch_assoc($result)) {
                             $age[] = $row;          

?>


                <br><br><br>





<input type="text" id="company_name" value="<?php  echo isset($row["company_name"])?$row["company_name"]:''; ?>" name="company_name" style="width: 460px;height: 50px;overflow: hidden;" placeholder="Company Name*">
                <br>
                <br><br><br>
                <input type="text" id="acess_code" value="<?php  echo isset($row["acess_code"])?$row["acess_code"]:''; ?>" name="acess_code" style="width: 460px;height: 50px;overflow: hidden;" placeholder="Acess Code*">
                <br><br><br><br>
                <input type="text" id="register_on" value="<?php  echo isset($row["register_on"])?$row["register_on"]:''; ?>" name="register_on" style="width: 460px;height: 50px;overflow: hidden;" placeholder="Register On*">
                <br><br><br><br>

                <button name="editcompany"  id="editcompany" type="submit" style="border:0;width:100px;margin-left: 45px;" >
                <img src="images/save.png" alt="">

                </button>
                <button type="submit" style="border:0;width:100px;margin-left: 75px;">
                <img src="images/cancel.png" alt="">
                </button>




                <?php
}
                   }

 }
mysqli_close($conn);
                ?>          
                </form> 

                </div>

Answer 1

You shouldnt use != comparisment to check if a string is empty. Use strlen() > 0 instead. Secondly just debug whats in the $_REQUEST['id'] to know what value you get when you leave it empty. Then you can make your check work with that.

A second possibilty is to always set a default value for your combobox