When I insert a row into my ms-sql server database, I have the following error:
Warning: Variable parameter 3 not passed by reference (prefaced with an
&). Variable parameters passed tosqlsrv_prepareorsqlsrv_queryshould be passed by reference, not by value. For more information, seesqlsrv_prepareorsqlsrv_queryin the API Reference section of the product documentation. inC:\...\ImmoToevoegen.phpon line 6
The same warning for parameter 1 and 2.
This is my php code:
<?php
if (!empty($_POST['omschrijving']) && !empty($_POST['woningtype']) && !empty($_POST['data'])){
$oms = $_POST['omschrijving'];
$type = $_POST['woningtype'];
$data = json_encode($_POST['data']);
$prep = sqlsrv_prepare($conn, "insert into Woning(omschrijving, data, typeid) values (?, ?, ?);", array($oms, $data, $type));
sqlsrv_execute($prep);
print "Woning is toegevoegd";
}
?>
<form action="<?php print $_SERVER['PHP_SELF']; ?>" method="POST">
<!--param 1 "omschrijving"-->
Omschrijving:
<textarea name="omschrijving" id="omschrijving" cols="30" rows="10></textarea>
<!--param 2 "data"-->
<div>
wc: <input type="checkbox" name="data[]" value="wc"/><br/>
cv: <input type="checkbox" name="data[]" value="cv"/><br/>
boiler: <input type="checkbox" name="data[]" value="boiler"/><br/>
</div>
<!--param 3 "woningtype"-->
<select name="woningtype" id="woningtype">
<?php
//code for load data from database
print "<option value='" . $rij["ID"] . "'>" . $rij["naam"] . "</option>";
?>
</select>
</form>
<?php
sqlsrv_close($conn);
?>
I use like I said php with a ms-sql server database.
What is wrong with my code?
Thanks
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