Cannot access offset of type string on string

Question

Please help me to solve this problem.

**

public function uploadPhoto($file)
    {
        if(!empty($file)){
            $fileTempPath=$file['tmp_name'];
            $fileName=$file['name'];
            $fileType=$file['type'];
            $fileNameCmps=explode('.',$fileName);
            $fileExtension=strtolower(end($fileNameCmps));
            $newFileName=md5(time().$fileName) . '.' .
            $fileExtension;
            $allowedExtn=["png","jpg","jpeg"];
            if(in_array($fileExtension,$allowedExtn))
            {
                $uploadFileDir=getcwd().'/uploads';
                $destFilePath=$uploadFileDir . $newFileName;
                if(move_uploaded_file($fileTempPath,$destFilePath)){
                    return $newFileName;
                }
            }
    
        }
    }**

this is how I send data to parameters

** if($conf == $_SESSION['randomNum']) {

                    //create object of user class 
                    require_once 'C:\xampp\htdocs\MEDICALCENTERPROJECT\app\libraries\User.php';
                    $obj = new User();
                    $obj->connect();
                     
                    $img = "";
                    if($pic == "")
                    {
                        $img = "null";
                    }
                    else{
                        $img = $obj->uploadPhoto($pic);

                    }**

I tried many times to understand the error but I could not do. Here the error is shown in line 111 its mean $fileTempPath=$file['tmp_name']; in this line. Please help me to solve this.

this is how I sent data using ajax

**$(document).ready(function() {

$("#success").hide();


$("#con").click(function() {

    let confirm = $("#ccode").val();
    let isTrue = "false";
    let username = $("#pusername").val();
    let email = $("#pemail").val();
    let mobile = $("#pmobile").val();
    let addrs = $("#paddress").val();
    let id = $("#pidno").val();
    let pswd = $("#ppassword").val();
    let c_pswd = $("#pcpassword").val();
    let pic = $("#puserphoto").val();


    let data = new FormData("#addformSign");
    $.ajax({
        url: "./Patials/sendEmailVerifyCode.php", //"./Patials/ajax.php",
        type: "POST",
        data: {
             conf: confirm,
            isTrue: isTrue,
            username: username,
            email: email,
            mobile: mobile,
            addrs: addrs,
            id: id,
            pswd: pswd,
             pic: pic,
         },

        success: function(response) {

            alert(response);

            if (response == confirm) {

                $("#emailCon").modal("show");
                $("#confirm").hide();
                $("#success").show();

            } else {

                $("#ccode").css({
                    "borderColor": "red",
                    "backgroundColor": "rgba(253, 127, 127, 0.5)"
                });
                $("#confirm").show();
                $("#errMsg").text("Invalid Code!");

            }



        },

        error: function(request, error) {
            console.log(arguments);
            console.log("Erro" + error);
            $("#confirm").show();
        },
    });


});

**

let confirm = $("#ccode").val(); this input field is not in the same form.

Answer 1

I think you are sending string as $pic value. It has to be some sort of array. Make sure you are sending the file correctly with Ajax. Check this Debug by examining the real-time value of the variable using PHP's var_dump function.

The way to solve errors without internet (without asking anyone) is to search for the real-time value of the variable. You can do this in php with the var_dump/var_export/echo/print functions.

Make sure you provide enough information to the other party when asking questions. I don't know how you send data with Ajax. It's impossible for me to get the same error as you. You should ask the same in a way that I can provide or imagine in my own setting.

The error you get is TypeError, that is, you can fix this error by checking the data you send and the data you receive.

---

UPDATE:

What you need to be careful about is that all the <input> elements you want to send are in the <form>.

you can try like that.

$("#con").click(function() {
  let confirm = $("#ccode").val();

  let yourFormData = new FormData("#addformSign"); // all sended element must be inside this form

  $.ajax({
    url: "./Patials/sendEmailVerifyCode.php", //"./Patials/ajax.php",
    type: "POST",
    data: yourFormData,
    success: function(response) {
      alert(response);
      if (response == confirm) {
        $("#emailCon").modal("show");
        $("#confirm").hide();
        $("#success").show();
      } else {
        $("#ccode").css({
            "borderColor": "red",
            "backgroundColor": "rgba(253, 127, 127, 0.5)"
        });
        $("#confirm").show();
        $("#errMsg").text("Invalid Code!");
      }
    },
    error: function(request, error) {
      console.log(arguments);
      console.log("Erro" + error);
      $("#confirm").show();
    },
  });
});

---

If the above code does not solve the problem, make sure that you send the values ​​of the input fields you sent correctly with ajax.

For example, you need to use the following line as

let pic = $('#puserphoto').prop('files')[0];

instead of

let pic = $("#puserphoto").val();

---

also $_FILES is used to receive the file sent with the POST method.

$pic = $_FILES['pic']

Good Luck