php符号显示

Question

我想用php和mysql显示某个货币符号，表名是fund，表行名是symbol。该符号必须根据该特定行的 id 动态更改。我将如何编写应用程序用户无法更改的代码？基本上符号出现并且无法编辑。

如果我需要显示更多代码，请告诉我。

<label>Price</label>
  Symbol must display here- example: ($)      
<input type="text" name="price" value="<?php value('price'); ?>" />

我的错误部分:

<?php
$errors = array();
  if(isset($_POST['submit'])) {    //Check for postback
    if($_POST['price'] == '')
       $errors['price'] = 'Please enter a price.';
    if($_POST['date'] == '')
       $errors['date'] = 'Please enter a date.';
    if(count($errors) == 0){
    //edit funds
    $sql = "UPDATE funds SET price='" . $_POST['price'] . 
    "', date='" . $_POST['date'] . "' WHERE id = " . $_GET['id'];
    query($sql);
    set_flashdata('success', 'Fund updated');
    redirect('index.php');
  }  
 } else {
      $id = $_GET['id'];
      $sql = 'SELECT * FROM funds WHERE id = ' . $id;
      $results = query($sql);
      $_POST['price'] = $results[0]['price'];
      $_POST['date'] = $results[0]['date'];
  }
?>

Answer 1

<?php

function chk_symbol($id)
{
    global $sym;
    $array_symbol=array("$","@","%","#");
    $c=count($array_symbol);
    $i=0;
    for($i;$i<$c;$i++)
    {
        if($id==$i)
        {
            $sym=$array_symbol[$i];

        }
    }   

}

//call function which set a value of variable $sym and echo it wherever you want
chk_symbol(0);
echo $sym;

?>