私はStringRequestを使用してユーザーをログインするアンドロイドアプリを持っています。最近、いくつかのPHPコードが変更されました。サーバーエラー500が返されます。(エラーの意味を知っています)残念ながら、PHPはエラーコード500を返します。エラーはどこですか?
私のPHPコード:
<?php
require("password.php");
$con = mysqli_connect("mywebsite.com", "username", "password", "dbname");
$username = $_POST["username"];
$password = $_POST["password"];
$statement = mysqli_prepare($con, "SELECT * FROM user WHERE username = ?");
mysqli_stmt_bind_param($statement, "s", $username);
mysqli_stmt_execute($statement);
mysqli_stmt_store_result($statement);
mysqli_stmt_bind_result($statement, $colUserID, $colName, $colUsername, $colTheme, $colEmail, $colDefaultRadius, $colPassword, $timeElapsed1, $timeElapsed2, $timeElapsed3, $timeElapsed4, $timeElapsed5, $timeElapsed6, $timeElapsed7);
$response = array();
$response["success"] = false;
while(mysqli_stmt_fetch($statement)){
if (password_verify($password, $colPassword)) {
$response["success"] = true;
$response["name"] = $colName;
$response["user_id"] = $colUserID;
$response["theme"] = $colTheme;
$response["email"] = $colEmail;
$response["radius"] = $colDefaultRadius;
$response["timeElapsed1"] = $timeElapsed1;
$response["timeElapsed2"] = $timeElapsed2;
$response["timeElapsed3"] = $timeElapsed3;
$response["timeElapsed4"] = $timeElapsed4;
$response["timeElapsed5"] = $timeElapsed5;
$response["timeElapsed6"] = $timeElapsed6;
$response["timeElapsed7"] = $timeElapsed7;
}
}
mysqli_stmt_close($statement);
$statement2 = mysqli_prepare($con, "SELECT * FROM location WHERE username = ?");
mysqli_stmt_bind_param($statement2, "s", $username);
mysqli_stmt_execute($statement2);
mysqli_stmt_store_result($statement2);
mysqli_stmt_bind_result($statement2, $colUsername, $1, $2, $3, $4, $5, $6, $7, $8, $9, $10, $11, $12, $13, $14, $15, $16, $17, $18, $19, $20, $21, $22, $23, $24, $25, $26, $27, $28, $29, $30, $31, $32, $33, $34, $35, $36, $37, $38, $39, $40, $41, $42, $43, $44, $45, $46, $47, $48, $49);
while(mysqli_stmt_fetch($statement2)) {
if($response["success"] == true) {
$response["1"] = $1;
$response["2"] = $2;
$response["3"] = $3;
$response["4"] = $4;
$response["5"] = $5;
$response["6"] = $6;
$response["7"] = $7;
$response["8"] = $8;
$response["9"] = $9;
$response["10"] = $10;
$response["11"] = $11;
$response["12"] = $12;
$response["13"] = $13;
$response["14"] = $14;
$response["15"] = $15;
$response["16"] = $16;
$response["17"] = $17;
$response["18"] = $18;
$response["19"] = $19;
$response["20"] = $20;
$response["21"] = $21;
$response["22"] = $22;
$response["23"] = $23;
$response["24"] = $24;
$response["25"] = $25;
$response["26"] = $26;
$response["27"] = $27;
$response["28"] = $28;
$response["29"] = $29;
$response["30"] = $30;
$response["31"] = $31;
$response["32"] = $32;
$response["33"] = $33;
$response["34"] = $34;
$response["35"] = $35;
$response["36"] = $36;
$response["37"] = $37;
$response["38"] = $38;
$response["39"] = $39;
$response["40"] = $40;
$response["41"] = $41;
$response["42"] = $42;
$response["43"] = $43;
$response["44"] = $44;
$response["45"] = $45;
$response["46"] = $46;
$response["47"] = $47;
$response["48"] = $48;
$response["49"] = $49;
}
}
echo json_encode($response);
?>
私はエラーが$statement2
声明の中で起こっている知っていることに注意してください。私はいくつかのことを試しましたが、エラーを修正することはできません。 $1
〜$49
はデータベースのvarcharsにリンクしています。誰が私に何が起こっているか教えてもらえますか?私はVolley Response.ErrorListenerを使用しようとしましたが、何も表示されません。助けをあらかじめありがとう!
私にお知らせいただきありがとうございます!これは私の問題を解決しました。 – TheAnonymous010
@ TheAnonymous010問題が解決した場合は、回答を受け入れられた回答(または鉱山xdですが、付与されていますが、これはより良い形式であり、ソースがあります): –
回答としてマークしました:) – TheAnonymous010