Load Data from a mySql DB in to HTML textBox

Question

I'm trying to load data from a database table within my page.

this that I report below is a demonstrative example, to explain how I would like to realize it ... actually we are talking about 100 fields of the database and 1000 and passes rows of code of the table ...

<?php
$servername = "localhost";
$username = "progettocantiere";
$password = "";
$dbname = "my_progettocantiere";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 



$idCantiere = $_GET['idCantiere'];

$sql1 = "SELECT * FROM Cantiere WHERE idCantiere = '$idCantiere'";



    $result = mysqli_query($sql1, $conn);
    
   
    $details = mysqli_fetch_array($result);




    $savedNomeCantiere = $details["nomeCantiere"];
    $savedCodiceCommessa = $details["codiceCommessa"];
    $savedIndirizzoCantiere = $details["indirizzoCantiere"];

var_dump($details);


$result1 = $conn->query($sql1);

echo($nomeCantiere);

?>


<html>
<head>
</head>
<body>


   
        <table width="300" border="0">
          <tr>
            <td>Name</td>
            <td><input type="text" name="savedNomeCantiere" style="text-align:right" value="<?php echo $savedNomeCantiere; ?>" /></td>
          </tr>
          <tr>
            <td>Cost</td>
            <td><input type="text" name="savedCodiceCommessa" style="text-align:right" value="<?php echo $savedCodiceCommessa; ?>" /></td>
          </tr>
           <tr>
            <td>Cost</td>
            <td><input type="text" name="savedIndirizzoCantiere" style="text-align:right" value="<?php echo $savedIndirizzoCantiere; ?>" /></td>
          </tr>
          
        </table>

<br/>

   
</body>
</html>

I tried with this type of upload that is putting an "echo" inside the "value" of the textbox, but it does not work .

This line is used to derive the "id" through page redirection.

$idCantiere = $_GET['idCantiere'];

if i want to try a var_dump($details) it return NULL

You cannot mix mysqli and mysql APIs. Switch everything to mysqli. — aynber, Oct 02 '18 at 13:15
Possible duplicate of [Can I mix MySQL APIs in PHP?](https://stackoverflow.com/questions/17498216/can-i-mix-mysql-apis-in-php) — aynber, Oct 02 '18 at 13:15
Your query and your fetch both use mysql_* instead of mysqli — aynber, Oct 02 '18 at 13:18
Use [mysqli_query](https://secure.php.net/manual/en/mysqli.query.php) and [mysqli_fetch_array](https://php.net/manual/en/mysqli-result.fetch-array.php) instead. I'd also suggest you use [prepared statements](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php) and [bind_param](http://php.net/manual/en/mysqli-stmt.bind-param.php) to [prevent SQL injection](https://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php). — aynber, Oct 02 '18 at 13:24
You have errors in your query like @aynber says. So Correct your error. — Sfili_81, Oct 02 '18 at 13:46
@aynber i've changed mysql_query to mysqli_query and mysql_fetch_array to mysqli_fetch_array .. but doesn't work :( — NarcosZTK_10, Oct 02 '18 at 13:59
Try checking for [mysqli errors](http://php.net/manual/en/mysqli.error.php) after each mysqli call, to see if it's working correctly or throwing er — aynber, Oct 02 '18 at 14:02
okay i've tryed but nothing... i think that it's working correctly... @aynber — NarcosZTK_10, Oct 02 '18 at 14:14
it's a problem of html and the way to pass the $nomeCantiere in to the textBox? @aynber — NarcosZTK_10, Oct 02 '18 at 14:19
Can you [edit] your post with the current code? Either it's still not working, or there are no results. — aynber, Oct 02 '18 at 14:25
Let us [continue this discussion in chat](https://chat.stackoverflow.com/rooms/181159/discussion-between-narcosztk-10-and-aynber). — NarcosZTK_10, Oct 02 '18 at 14:26
`$result = mysqli_query($sql1, $conn);` should be `$result = mysqli_query($conn, $sql1);` — waterloomatt, Oct 02 '18 at 15:34

khalid · Accepted Answer · 2018-10-02T15:23:19.997

your code shoud be like this

<?php
$servername = "localhost";
$username = "progettocantiere";
$password = "";
$dbname = "my_progettocantiere";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 



$idCantiere = $_GET['idCantiere'];

$sql1 = "SELECT * FROM Cantiere WHERE idCantiere = '{$idCantiere'}";



    $result = $conn->query($sql1);


    $details = $result->fetch_array();




    $savedNomeCantiere = $details["nomeCantiere"];
    $savedCodiceCommessa = $details["codiceCommessa"];
    $savedIndirizzoCantiere = $details["indirizzoCantiere"];

var_dump($details);


$result1 = $conn->query($sql1);

echo($nomeCantiere);

?>


<html>
<head>
</head>
<body>



        <table width="300" border="0">
        <tr>
            <td>Name</td>
            <td><input type="text" name="savedNomeCantiere" style="text-align:right" value="<?php echo $savedNomeCantiere; ?>" /></td>
        </tr>
        <tr>
            <td>Cost</td>
            <td><input type="text" name="savedCodiceCommessa" style="text-align:right" value="<?php echo $savedCodiceCommessa; ?>" /></td>
        </tr>
        <tr>
            <td>Cost</td>
            <td><input type="text" name="savedIndirizzoCantiere" style="text-align:right" value="<?php echo $savedIndirizzoCantiere; ?>" /></td>
        </tr>

        </table>

<br/>


</body>
</html>

Don't just post a code dump. Please explain why this solution works. — waterloomatt, Oct 02 '18 at 15:38
than this code $result = mysqli_query($sql1, $conn); shoud be $result = $conn->query($sql1); and this $details = mysqli_fetch_array($result); shoud be $details = $result->fetch_array(); because they used $conn = new mysqli to connect to db, you can use the original code of "$result = mysqli_query($sql1, $conn); $details = mysqli_fetch_array($result);" if you open connection with db like that $conn = mysqli_connect(...); ;) — khalid, Oct 02 '18 at 17:19

score 0 · Answer 2 · answered Oct 02 '18 at 15:53

One of the issues is that you're passing $conn and $sql1 in the wrong order.

$result = mysqli_query($sql1, $conn); should be $result = mysqli_query($conn, $sql1);

if i want to try a var_dump($details) it return NULL

That is exactly why you need to check $result before using it. mysqli_query returns an object if it was successful or false if it failed - http://php.net/manual/en/mysqli.query.php#refsect1-mysqli.query-returnvalues.

You don't need to get fancy here but you should at least do a sanity check.

...
$sql1 = "SELECT * FROM Cantiere WHERE idCantiere = '$idCantiere'";
$result = mysqli_query($sql1, $conn);

if ($result !== false) {
    $details = mysqli_fetch_array($result);
    ...

Also, I'm required to point out "SELECT * FROM Cantiere WHERE idCantiere = '$idCantiere'"; is very dangerous because you don't have control over $idCantiere. Please lookup SQL Injection and how to avoid it.

Load Data from a mySql DB in to HTML textBox

2 Answers2