Problem to update record with php and mysql combobox?

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Problem to update record with php and mysql combobox?

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#1 by (1 votes)
#2 by (0 votes)

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I'm doing the update of a record in my BD by <select> but it does not work for me.

monitoringService.php

<html>
<head>
  <form method="POST" action="actualizarSeguimiento.php">
    <label>Situacion: </label>
        <select type="selec" name="situacion[]" id="situacion">
            <option value='".$situacion."'>Finalizado</option>
            <option value='".$situacion."'>Fallido</option>
            <option value='".$situacion."'>Cancelado</option>
            <option value='".$situacion."'>Programado</option>
            <option value='".$situacion."'>Realizando</option>
        </select>
   </form>    
</body>
</html>

updateTracking.php

<?php
    $conexion =  mysqli_connect("localhost", "root", "admin123", "database");

    $servicio_No = $_POST["servicio_No"];
    $situacion = $_POST["situacion"];

    mysqli_query($conexion,"UPDATE servicio SET situacion = '$situacion'   WHERE servicio_No = '$servicio_No'") or die("ERROR AL ACTUALIZAR");

   mysqli_close($conexion);
?>

mysql php

asked by Cristian Antonio Trujillo Gris 08.06.2018 в 22:02

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2 answers

1

I do not understand what you want to do, but you will need something similar to this, considering that you need to set the variable $servicio_No previously.

monitoringService.php

<html>
    <head></head>
    <body>
        <form method="POST" action="actualizarSeguimiento.php">
            <input type="hidden" name="servicio_No" value="<?php echo $servicio_No ?>" />

            <label>Situacion: </label>
            <select name="situacion" id="situacion" onchange="this.form.submit()">
                <option value="Finalizado">Finalizado</option>
                <option value="Fallido">Fallido</option>
                <option value="Cancelado">Cancelado</option>
                <option value="Programado">Programado</option>
                <option value="Realizando">Realizando</option>
            </select>
       </form>    
    </body>
</html>

You need to set servicio_No , since that is what you get to do the update in your query, also assuming that the field situacion of the table servicio is alphanumeric

updateTracking.php

<?php
    $conexion =  mysqli_connect("localhost", "root", "admin123", "database");

    $servicio_No = $_POST["servicio_No"];
    $situacion = $_POST["situacion"];

    mysqli_query($conexion,"UPDATE servicio SET situacion = '$situacion'   WHERE servicio_No = '$servicio_No'") or die("ERROR AL ACTUALIZAR");

    mysqli_close($conexion);
?>

answered by 08.06.2018 в 22:07

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score 0 · Accepted Answer

Your <head> tag is not closed and your </body> tag is unopened.

<html>
<head></head>
<body>
  <form method="POST" action="actualizarSeguimiento.php">
    <!-- Código de tu formulario -->
   </form>    
</body>
</html>

As you define the action of your form, your updateFollowing.php must be in the same folder as trackingService.php . If it is in another folder you should specify it.
I would remove the brackets from the attribute name of select - > name="situacion" unless you are looking for a multiple selection.
Is the variable $situacion of your <option> where it comes from?
Where servicio_No comes from, which you recover by POST in updateTracking.php - > $servicio_No = $_POST["servicio_No"];
Why do all the options pass the same variable to the value attribute? I think the idea would be that the value would be different depending on which option they choose.
I recommend that to retrieve values with $ _POST use the following function $situacion = mysqli_real_escape_string($conexion, $_POST['situacion']); You can find more information about this function here .
If you are not sure if your values are going to update tracking correctly, you can add a echo $situacion to the end of that same file to see what value the variable has and be able to debug errors.