insert multiple input text at the same name in sql and php

Question

i want to insert multiple input in the same table using php and sql

example:

<script language="javascript">
fields = 0;
function addInput() {
if (fields != 10) {
document.getElementById('text').innerHTML += "<input type='text' name='order'><br/>";
fields += 1;
} else {
document.getElementById('text').innerHTML += "<br />Only 10 insert fields allowed.";
document.form.add.disabled=true;
}
}
</script>


<form action='insert.php' method='post'>
<input type='text' name='order'>


<div id="text">

</div>

<br /><input type='button' value='add new input' onclick='addInput()'> <input type='submit' value='submit'>

</form>

in this code i have unlimited input, i have to insert it in:

dbtable = clients_order

row name : order

++++++++++++++++++
+  id  +  order  +
++++++++++++++++++

so when i use this code :

$order=$_POST['order'];
foreach ($order as $insert_order) {

mysql_query("INSERT INTO  clients_order (order)
    VALUES ('$insert_order')");

 }

Error message : Warning: Invalid argument supplied for foreach()

please help me to do that

thank you

staticsan · Accepted Answer · 2011-10-05 02:45:33Z

2

The foreach() is complaining that $order isn't an array. If you use order[] as the name field in the HTML for your field, then PHP will turn them all into an array for you.

You should also check you have an array (use is_array()) before giving it to a foreach() loop. And you should call mysql_real_escape_string() over each piece of text from $orders in order to prevent SQL injection attacks.

answered Oct 5, 2011 at 2:45

staticsan

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2 Comments

Ali Khatem Over a year ago

good, now how can i insert 2 input in one foreach function example : foreach ($order,$another as $insert_order,$another) {

staticsan Over a year ago

Ideally, you need a generator, but PHP doesn't have those (though there are ways to do it anyway). Instead, you need to look over each array separate and re-assemble the data in another array. Then you loop over that to construct your SQL.

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