Arrayed input value inserts empty record in mysql

Question

I use 4 input boxes with same name.I will fill only one input box at a time and insert in DB using php.But doing following method inserts three empty records along with value-filled 4 th record.Need help!

Html

                <input type="text" name="abc[]" id="abc" value=""> 
                <input type="text" name="abc[]" id="abc"  value="">
                <input type="text" name="abc[]" id="abc"  value="">
                <input type="text" name="abc[]" id="abc" value="">

Php

$abc = $_POST['abc'];
        if (is_array($abc)) {
                foreach($abc as $c) {

$insert="INSERT INTO item(productid) VALUES ('$c')";
                    $insert2=mysql_query($insert);              }

}

Reza-S4 · Accepted Answer · 2014-12-14 21:08:48Z

3

just check if the value is not empty then insert to database.

$abc = $_POST['abc'];
if (is_array($abc)) {
    foreach ($abc as $c) {
        if(!empty($c)){
            $insert = "INSERT INTO item(productid) VALUES ('$c')";
            $insert2 = mysql_query($insert);
        }
    }

}

answered Dec 14, 2014 at 21:08

Reza-S4

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Comments

david strachan · Accepted Answer · 2014-12-14 21:47:14Z

1

The following code produces 1 INSERT statement. If as you suggest you are only using 1 value from your input, and can enforce this you can omit $value = rtrim($value, ","); which removes the trailing ,.

I would also advise you to use either mysqli or PDO because mysql_functions are deprecated.

 $value = "";
if(isset($_POST['submit'])){
    if(isset($_POST['abc'])){
        foreach($_POST['abc'] as $c) {
        if(!empty($c)){
        $value .= "'".$c."',";}
        }
        $value = rtrim($value, ",");
    }
    $insert="INSERT INTO item(productid) VALUES ($value)";
    echo $insert;//Remove after testing
}

answered Dec 14, 2014 at 21:47

david strachan

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