MySQL database entries only inserting the last user input from multiple form input clones

Question

Example of the SQL/php code, the second query is applicable. I have created a user input that submits the id variables into a MySQL database. These user inputs are cloned multiple times within the form, thus resulting in multiple duplicate id's/names, however one column (lets call it column 1) is different for each clone.

Now, once the input has been added and the form is submitted it only appends the last form entry into the database table, which I expected.

Is there a way that all these copies can be added to the table, perhaps based on the variance of Column 1? I was thinking something like: for each Column 1 INSERT INTO database1 ...

EDIT

Some code:

HTML:

<style> 
.hidden {
    display: none;
}
</style>

<body>
    <div id="samplesinfo" class="samplesinfo hidden">
        <input type="text" id="column1" name="column1" value="1" readonly>
        <input type="text" name="column2">
    </div>
    <button id="btn">Paste</button>
    <div>
        <h3>Paste Below</h3>
    </div>
    <form>
        <div id="paste">    
        </div>
        <button type="submit" name="submit">Submit</button> 
    </form>

jQuery:

$(document).ready(function() {
        var sample = 1;
        $("#btn").click(function() {
            var element = $(".samplesinfo.hidden").clone(true);
            element.removeClass("hidden").appendTo("#paste:last");
            sample++;
            $("#column1").val(sample)
        });

    });

and php:

if (isset($_POST['submit'])) {  
    $column1 = mysqli_real_escape_string($conn, $_POST['column1']);
    $column2 = mysqli_real_escape_string($conn, $_POST['column2']);

    $query = "INSERT INTO db1 ('column1', 'column2')
    VALUES ('$column1', '$column2');";

    mysqli_query($conn, $query);
    header("Location: ../web_page/analysis_page.php?add_order=success");
}

How does your form look ? In general, you should give every input it's own name, so you can tell the difference when you get the posted information. If there's no sensible difference, just enumerate them like name1, name2, etc and then have your storage script enumerate them the same way. — Gloweye
– Gloweye, Commented Jun 21, 2018 at 9:31
@JaccovanDorp the problem is that the clones are based on user preference. So the user would select the amount of clones needed. — Mariska
– Mariska, Commented Jun 21, 2018 at 9:33
@samezedi I would love to show you the code, but then I would have to attach the entire project which is quite extensive. — Mariska
– Mariska, Commented Jun 21, 2018 at 9:34
maybe you could just do a screen shot of the code which inserts into the db table — samezedi
– samezedi, Commented Jun 21, 2018 at 9:37

rkeet · Accepted Answer · 2018-06-21 11:11:04Z

1

That only the new inputs are send to the back-end is because the original #column1 and #column2 are outside of the <form>. If you want to send everything, put it in the <form>.

Next, if you want to insert everything into the database and simultaneously update existing records if they're changed you might want to read this.

If you want to add multiple records of the same object, you must send them to the back-end as the same type of records and then treat them as an array.

<form>
    <input name="column1[0][name]" value="first one">
    <input name="column1[1][name]" value="second one">
    <input name="column1[2][name]" value="third one">
</form>

Submitting this example would get you 3 column1 array entries with a name and associated values.

// example value of received data
$_POST = [
    'column1' => [
        0 => [
            'name' => 'first one',
        ],
        1 => [
            'name' => 'second one',
        ],
        2 => [
            'name' => 'third one',
        ],
    ],
];

To dump all into database use your query in a foreach() loop. Such as:

foreach ($_POST['column1'] as $column) {
     $query = "INSERT INTO db1 ('column1') VALUES (" . $column['name'] . ");";
}

Obviously you'd have to update it a bit here 'n' there for your use-case and don't forget to sanitize this data.

edited Jun 21, 2018 at 11:11

answered Jun 21, 2018 at 10:38

rkeet

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7 Comments

Mariska Over a year ago

I am using your last solution just to dump everything but I am getting this error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting '-' or identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING)

rkeet Over a year ago

Contained a small syntax error in the sql, fixed that for you. As an added challenge for you: when adding multiple records at once, do it in a single query. Added on top of that: if more than 100 records per query, split into multiple queries and insert via transactions. Just to keep it a challenge ;)

Mariska Over a year ago

I am guessing you use the mysqli_multi_query function for multiple queries :). Just a question regarding your sql, do I have to add column2 info as you would normally do after INSERT INTO and then would VALUES be like this: (" . $column['column1'] . $column['column2']");

Mariska Over a year ago

I think in your example you are only using one column, how about two or more?

rkeet Over a year ago

Yes I created a limited example. Found that understanding is aided more when people still have to figure something out :) On that note, yes, the example is just 1 column. If you want column 2 added, add it in the form to be send with it's own input (e.g.: <input name="column2[0][name]" value="test value">). This would create an addition top-level array in your $_POST data ($_POST['column1' => [... its data in answer], 'column2' => [ // column2 data]]).

|

Adder · Accepted Answer · 2018-06-21 14:19:17Z

0

I replaced the ids with classes and added a [] to the input field names, so that $_POST['column1'] and $_POST['column2'] are arrays.

HTML:

<style> 
.hidden {
    display: none;
}
</style>

<body>
    <div id="samplesinfo" class="samplesinfo hidden">
        <input type="text" class="column1" name="column1[]" value="1" readonly>
        <input type="text" class="column2" name="column2[]">
    </div>
    <button class="btn">Paste</button>
    <div>
        <h3>Paste Below</h3>
    </div>
    <form>
        <div class="paste">    
        </div>
        <button type="submit" name="submit">Submit</button> 
    </form>

Javascript:

$(document).ready(function() {
        var sample = 1;
        $(".btn").click(function() {
            sample++;
            $(".samplesinfo.hidden").find(".column1").val(sample);
            var element = $(".samplesinfo.hidden").clone(true);
            element.removeClass("hidden").appendTo(".paste:last");
        });

    });

PHP:

if (isset($_POST['submit'])) {  
    $n = count($_POST['column1']);
    for($i = 0; $i<$n; ++$i) {
        $column1 = mysqli_real_escape_string($conn, $_POST['column1'][$i]);
        $column2 = mysqli_real_escape_string($conn, $_POST['column2'][$i]);

        $query = "INSERT INTO db1 (column1, column2)
        VALUES ('$column1', '$column2');";

        mysqli_query($conn, $query);
    }
    header("Location: ../web_page/analysis_page.php?add_order=success");
}

edited Jun 21, 2018 at 14:19

answered Jun 21, 2018 at 10:36

Adder

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5 Comments

Mariska Over a year ago

Why did you change the ids to classes?

Adder Over a year ago

I changed the ids to classes because ids are supposed to be unique, while classes are not unique. Duplicate ids could cause problems. For example, your code $("#column1").val(sample) will change all ids to sample-value, not just the last one.

Mariska Over a year ago

I am not sure if I am doing something wrong, but it results in single entry and the entries are single letters.

Adder Over a year ago

Did you change the name="column1[]" and name="column2[]" to have brackets []?

Mariska Over a year ago

I did but it now only results in the last input appended in the database. Please also note that the ' ' should be removed from the INSERT INTO variables, this did the trick, thank you!

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