php - How to save multiple data from dynamically added text-boxes (JavaScript) into MySQL database? - Stack Overflow

admin管理员组

文章数量:1420228

I have been trying for this but no success. I am using PHP, HTML, JavaScript, and MySQL. Here is my HTML code:

<div id="Author">
    <LI>Author</LI> 
    <input type = "text" name="Author[]" value = "1. "/>
    <input type="button" value="+" id="Authorbutton" onclick="addAuthor()" />
</div>

If add button is clicked, another textbox will appear and user and put in another name. Here is my JavaScript:

var counter3 = 0;
function addAuthor() {
    // Get the main Div in which all the other divs will be added
    var mainContainer = document.getElementById('Author');

    // Create a new div for holding text and button input elements
    var newDiv = document.createElement('div');

    // Create a new text input
    var newText = document.createElement('input');
    newText.type = "text";
    //var i = 1;
    newText.name = "Author[]";
    newText.value = counter3 + 2 + ". ";

    //Counter starts from 2 since we already have one item
    //newText.class = "input.text";
    // Create a new button input
    var newDelButton = document.createElement('input');
    newDelButton.type = "button";
    newDelButton.value = "-";

    // Append new text input to the newDiv
    newDiv.appendChild(newText);
    // Append new button input to the newDiv
    newDiv.appendChild(newDelButton);
    // Append newDiv input to the mainContainer div
    mainContainer.appendChild(newDiv);
    counter3++;
    //i++;

    // Add a handler to button for deleting the newDiv from the mainContainer
    newDelButton.onclick = function() {
        mainContainer.removeChild(newDiv);
        counter3--;
    }
}

Here is my PHP code:

if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db($database, $con);



$sql="INSERT INTO savetest (type, number)
VALUES
('$_POST[type]','$_POST[Author]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";

mysql_close($con);
echo "Thank you for submitting your details!";

I heard a lot of people can get this to work by using arrays, but the data I stored in the database is Array. It does not matter how many textboxes I've created, just Array.

Am I using the correct approach? Should I save this array of data in one database fields?

I have been trying for this but no success. I am using PHP, HTML, JavaScript, and MySQL. Here is my HTML code:

<div id="Author">
    <LI>Author</LI> 
    <input type = "text" name="Author[]" value = "1. "/>
    <input type="button" value="+" id="Authorbutton" onclick="addAuthor()" />
</div>

If add button is clicked, another textbox will appear and user and put in another name. Here is my JavaScript:

var counter3 = 0;
function addAuthor() {
    // Get the main Div in which all the other divs will be added
    var mainContainer = document.getElementById('Author');

    // Create a new div for holding text and button input elements
    var newDiv = document.createElement('div');

    // Create a new text input
    var newText = document.createElement('input');
    newText.type = "text";
    //var i = 1;
    newText.name = "Author[]";
    newText.value = counter3 + 2 + ". ";

    //Counter starts from 2 since we already have one item
    //newText.class = "input.text";
    // Create a new button input
    var newDelButton = document.createElement('input');
    newDelButton.type = "button";
    newDelButton.value = "-";

    // Append new text input to the newDiv
    newDiv.appendChild(newText);
    // Append new button input to the newDiv
    newDiv.appendChild(newDelButton);
    // Append newDiv input to the mainContainer div
    mainContainer.appendChild(newDiv);
    counter3++;
    //i++;

    // Add a handler to button for deleting the newDiv from the mainContainer
    newDelButton.onclick = function() {
        mainContainer.removeChild(newDiv);
        counter3--;
    }
}

Here is my PHP code:

if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db($database, $con);



$sql="INSERT INTO savetest (type, number)
VALUES
('$_POST[type]','$_POST[Author]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";

mysql_close($con);
echo "Thank you for submitting your details!";

I heard a lot of people can get this to work by using arrays, but the data I stored in the database is Array. It does not matter how many textboxes I've created, just Array.

Am I using the correct approach? Should I save this array of data in one database fields?

• php
• javascript
• mysql
• html

Share Improve this question Follow edited Apr 20, 2012 at 9:20 Mat 207k4141 gold badges402402 silver badges418418 bronze badges asked Apr 19, 2012 at 20:10 GeorgioCZYGeorgioCZY 2311 silver badge33 bronze badges 4

• GeorgioCZY, I've read this over three times and I still can't figure out what you are asking. Do you want another text field to appear? Is it not appearing? Also, your code has several formatting issues. Please revise and it will be much easier to help. – Brendon Cheves Commented Apr 19, 2012 at 22:52
• if you want to save all data to one single field you need to implode data implode(',', $_POST[Author]). – Nisanth Kumar Commented Apr 20, 2012 at 9:50
• stackoverflow./questions/1978438/save-php-array-to-mysql – Nisanth Kumar Commented Apr 20, 2012 at 9:55
• @BrendonCheves My javascript adding textbox already worked, I just want to save all input into Mysql. right now, it only saves the first input. If i use array, only text "array" will be saved. My question is how to save all these data into mysql. I am not sure if I should save all these textbox data into one single field. – GeorgioCZY Commented Apr 20, 2012 at 13:33

Add a ment  | 

1 Answer 1

Sorted by: Reset to default 3

if (!$con){
   die('Could not connect: ' . mysql_error());
}

mysql_select_db($database, $con);

$authors = $_POST['Author'];
foreach($authors as $author) :
    $sql="INSERT INTO savetest (type, number) VALUES ('{$_POST['type']}','{$author}')";

    if (!mysql_query($sql,$con)){
       die('Error: ' . mysql_error());
    }
endforeach;

本文标签：