将数组转换为字符串或对象

Question

Good day, I am trying to convert an array "$list" into string or object. I have used following methods:

<?php

include "medclass.php";

session_start();
if (isset($_SESSION['mail'])) 
{
    $list = $_SESSION['basket'];
}
else
header("location: clientsigninpage.php?msg= Log-in First");

$obj = new med_class;
$obj->connectdb();

$val = implode(";",$list);    //implode method
$val = (object) $list;        //object method
$val = serialize($list);      //serialize method

$result = $obj->searchMed($val);

while ($row = $result->fetchObject()) 
{
  echo $row->MedPrice;
}

?>

With "(object)" its giving me following error: "Object of class stdClass could not be converted to string", with "implode": "Array to string conversion" and with "serialize()" it does not print anything.

The function that I am passing value is:

function searchMed($v1)
    {
        $sql = "select * from storepreview where MedName = '$v1'";
        $ret = $this->con->query($sql);
        return $ret;
    }

I have used these methods by seen following links: ( http://www.dyn-web.com/php/arrays/convert.php ) ; ( Convert an array to a string ); ( How to convert an array to object in PHP? )

Answer 1

You can use json_encode to convert Array to String:

$FINAL_VALUE = json_encode($YOUR_OBJECT);

For more information, you can refer this link .

Answer 2

I managed to reproduce your "Array to string conversion" error when using the implode command by running the following line of code:

implode(";", [[]]); // PHP Notice:  Array to string conversion in php shell code on line 1

For converting a nested array into a string I found that a foreach loop worked:

$nestedArray = ['outerKeyOne' => ['innerKeyOne' => 'valueOne'], 'outerKeyTwo' => ['innerKeyTwo' => 'valueTwo']];
$arrayOfStrings = [];
foreach ($nestedArray as $key => $value) {
    $arrayOfStrings[] = implode(",", $value);
}
implode(";", $arrayOfStrings); // string(17) "valueOne;valueTwo"

The second error associated with the line $val = (object) $list; is from trying to embed an object into the $sql string. It seems like an object is not what you want here, unless it is an object that has a __toString() method implemented.

I hope this is of some help. Using var_dump or something similar would provide more debug output to better diagnose the problems along with the above error messages. That's how I came up with the above code.