У меня MySQL-запрос успешно обновил один из двух столбцов, но не другой. Успешный столбец, productionstage, имеет тип данных: varchar и неуспешный столбец, floornotes, имеет тип данных: текст. Я тестировал создание типа данных floornotes: varchar без успеха. Данные для floornotes и productionstage исходят из html-формы с ее данными, отправленными через ajax на php. productionstage – это выбор меню, а floornotes – текстовое поле. Есть идеи по этому вопросу? Я не получаю никаких ошибок с сервера, и когда я делаю console.log () в моем ajax для переданного значения floornotes, я вижу правильную строку каждый раз.
МОЙ КОД:
HTML:
<form id="workorderMovement" name='workorderMovement_form' action="workordermovementGET.php" method="post"> <fieldset id="userid"> <span>Welcome <?php echo $user ?> </span> </fieldset> <fieldset id="sgnum"> <fieldset id="fieldset" style="text-align: center;"> <span>Please enter the SG Number</span> </fieldset> <input type="text" name="sgnumber" id="sgnumber"> <input type="button" name="searchButton" id="searchButton" value="SEARCH"> </fieldset> <br/> <br/> <fieldset id="stageSelectField"> <fieldset id="fieldset" style="text-align: center;"> <span>Please select the Stage Completed</span> </fieldset> <select name="stageSelect" id="stageSelect"> <option value="Please Select">Please Select</option> <option value="Film Done">Film Done</option> <option value="Staged Done">Staged Done</option> <option value="Cleanroom Done">Cleanroom Done</option> <option value="GB2 Done">GB2 Done</option> <option value="Bagging Done">Bagging Done</option> <option value="Inspection Done">Inspection Done</option> <option value="LC Inspection Done">LC Inspection Done</option> <option value="IGU Done">IGU Done</option> </select> </fieldset> <br/> <br/> <fieldset id="floorNotesField"> <fieldset id="fieldset" style="text-align: center;"> <span>Please enter any new work order notes</span> </fieldset> <textarea type="text" name="floorNotes" id="floorNotes" class="floorNotesText"></textarea> </fieldset> <br/> <br/> <br/> </form> <!-- End Work Order Movement Form --> <fieldset id="doneButtonField"> <input type="button" name="doneButton" id="doneButton" value="DONE"> </fieldset> MY AJAX:
j("#doneButton").click(function(){ //send Workorder Movement Data values to php using ajax. var sgnumber = j('#sgnumber').val(); var stageselect = j('#stageSelect').val(); var floornotes = j('#floorNotes').val(); console.log(floornotes); j.ajax ({ method: 'POST', url: "workordermovementUPDATE.php", data: {sgNumber: sgnumber, stageSelect: stageselect, floorNotes: floornotes}, dataType: 'json', success: function(){} }); });
Мой PHP:
<?php include('inc.php'); //Get Table Options. if (isset($_POST['sgNumber'])) { $sgNumber = $_POST['sgNumber']; if (isset($_POST['stageSelect'])) { $stageSelect=$_POST['stageSelect']; } if (isset($_POST['floorNotes'])) { $floorNotes=$_POST['floorNotes']; } //connect to the database $conn = new mysqli($servername, $username, $password, $dbname); // Check connection if(mysqli_connect_errno() ) { printf('Could not connect: ' . mysqli_connect_error()); exit(); } $conn->select_db($dbname); if(! $conn->select_db($dbname) ) { echo 'Could not select database. '.'<BR>'; } $sql= "UPDATE invoicesDec2016 SET productionstage = ?, floornotes = ? WHERE id = ?"; $stmt = $conn->prepare($sql); $stmt->bind_param('ssi', $stageSelect, $floornotes, $sgNumber); $stmt->execute(); if(mysqli_query($conn, $stmt)){ echo "".$sgnumber." Updated Successfully!"; } else { echo "ERROR: Could not update ".$sgnumber."".mysqli_error($conn).""; } //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// //Free the result variable. $result->free(); //Close the Database connection. $conn->close(); }//End If statement ?>в<?php include('inc.php'); //Get Table Options. if (isset($_POST['sgNumber'])) { $sgNumber = $_POST['sgNumber']; if (isset($_POST['stageSelect'])) { $stageSelect=$_POST['stageSelect']; } if (isset($_POST['floorNotes'])) { $floorNotes=$_POST['floorNotes']; } //connect to the database $conn = new mysqli($servername, $username, $password, $dbname); // Check connection if(mysqli_connect_errno() ) { printf('Could not connect: ' . mysqli_connect_error()); exit(); } $conn->select_db($dbname); if(! $conn->select_db($dbname) ) { echo 'Could not select database. '.'<BR>'; } $sql= "UPDATE invoicesDec2016 SET productionstage = ?, floornotes = ? WHERE id = ?"; $stmt = $conn->prepare($sql); $stmt->bind_param('ssi', $stageSelect, $floornotes, $sgNumber); $stmt->execute(); if(mysqli_query($conn, $stmt)){ echo "".$sgnumber." Updated Successfully!"; } else { echo "ERROR: Could not update ".$sgnumber."".mysqli_error($conn).""; } //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// //Free the result variable. $result->free(); //Close the Database connection. $conn->close(); }//End If statement ?>
ПРИМЕЧАНИЕ. Когда я делаю обновление, успешно удаляются обновления productionstage и удаляется предыдущий контент floornotes.