Hp Parse Error Syntax Error Unexpected T_else
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T_else Php
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T_else Error In Php
each other. Join them; it only takes a minute: Sign up Parse error: syntax error, unexpected 'else' (T_ELSE) [closed] up vote -4 down vote favorite I edit this code to display a mesaje when select is null and i get this error: Parse error: syntax error, unexpected 'else' (T_ELSE) in /nginx/user/reports.php on line 235 This is code: $stmt unexpected end of file php = $mysqli->prepare("SELECT date, impressions, balance, username FROM reports WHERE username = '$username' and date between '$firstDay' AND '$lastDay'"); $stmt->execute(); $stmt->store_result(); $rows = $stmt->num_rows; mysqli_stmt_bind_result($stmt,$date,$impressions,$balance,$username); if ($rows > 0){ while (mysqli_stmt_fetch($stmt)) { $ecpm_t3 = $balance*1000; $ecpm3 = $ecpm_t3/$impressions; $rate = number_format((float)$ecpm3, 2, '.', ''); echo"
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Syntax Error, Unexpected 'else' (t_else) Laravel
more about Stack Overflow the company Business Learn more about hiring developers or posting php else ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack parse error: syntax error, unexpected $end Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up Parse error: syntax error, unexpected T_ELSE up vote 0 down vote http://stackoverflow.com/questions/21075325/parse-error-syntax-error-unexpected-else-t-else favorite i just need help regarding my code becuase i odnt have any idea where is the syntax error. $file = $_FILES['image']['tmp_name']; if (!isset($file)) echo "Please select an image"; else { $image = addslashes(file_get_contents($_FILES['image']['tmp_name'])); $img_name = addslashes($_FILES['image']['name']); $img_size = getimagesize($_FILES['image']['tmp_name']); if ($img_size == FALSE) echo "select valid image."; else { if (!$insert = mysql_query("insert into testblob values('','$image','$img_name')")); echo "Problem uploading!"; else { $lastid = mysql_insert_id(); echo "Image uploaded.
http://stackoverflow.com/questions/22562818/parse-error-syntax-error-unexpected-t-else />Your image:
"; } } } i just need to know where is the error. thnaks for the help. more power .. php mysql share|improve this question edited Mar 21 '14 at 15:28 Pavel Štěrba 1,6331629 asked Mar 21 '14 at 15:24 user2699948 816 Which line PHP reports you? –Pavel Štěrba Mar 21 '14 at 15:25 Indenting your code may make the solution obvious. –fedorqui Mar 21 '14 at 15:25 4 i know its possible to do if..else without curly braces, but you really should just always use curly braces even if its only one line... –celeriko Mar 21 '14 at 15:27 also, if (!$insert=mysql_query("insert into testblob values('','$image','$img_name')")); should not have a ; at the end.. –celeriko Mar 21 '14 at 15:27 @celeriko, found my error. thanks! –user2699948 Mar 21 '14 at 15:31 add a comment| 3 Answers 3 active oldest votes up vote 1 down vote You have a semi-colon ; in your conditional statement extra here: if (!$insert=mysql_query("insert into testblob values('','$image','$img_name')")); Remove it and change it to: if (!$insert=mysql_query("insert into testblob values('','$image','$img_name')")) why dont you use this better ? $file= $_FILES['image']['tmp_name']; if (!isset($file)){ echo "Please select an image"; } else{ $image= addslashes(file_get_contents($_FILES['PHP errors. These questions often pertain to very basic errors. Even with the error message, these users still need help. I have developed with PHP for over a decade. During that time I’ve encountered nearly every error. This post covers how http://jason.pureconcepts.net/2013/05/fixing-php-errors/ to interpret a PHP error as well as fixing common PHP errors. We will parse the https://php1st.com/578/ following PHP code and resolve the errors. There are three. Four depending on how you define errors (more on that later). For now, bonus points if you can find the other error. name) { echo 'It's time to stop writting errors "; echo $user->name, '!'; Interpreting PHP errors When we run this code, we receive the first error: syntax error PHP Parse error: parse error, expecting ‘,’ or ‘;’ in errors.php on line 3 Before we fix this error, let’s interpret the error. PHP errors have a three important parts: Error type found at the beginning tells us the error type. PHP has many error types. Parse or Fatal errors being more common. Example: PHP Parse error Error message provides us a hint about the error. Example: expecting ‘,’ or ‘;’ Error context tells us where the error occured. Example: parse error syntax errors.php on line 3 Together these parts provide all the information we need to fix our code. PHP Error #1: Expecting ‘,’ or ‘;’ PHP Parse error: parse error, expecting ‘,’ or ‘;’ in errors.php on line 3 The error tells us we have a parse error on line 3. Looking at line 3 again: if ($user->name) { Seems correct. What’s wrong? This is where error type can help solve the mystery. For parse errors, the error typically occurs on the preceeding line since the parser continues until it reads invalid syntax. Let’s look at line 2: echo 'Hello Errors!' Now if you wrote this code, you may not see the error. In which case the error message provides a hint: expecting ‘,’ or ‘;’. Expecting a comma… What? echo allows you to output multiple strings separated by commas. However, this was not our intention. Expecting a semi-colon… Ahh. The line is missing its required semi-colon line ending. Let’s fix the error by adding a semi-colon to the end of line 2. PHP Error #2: Unexpected T_STRING PHP Parse error: unexpected T_STRING in errors.php on line 4 Another parse error. Applying what we’ve learned, we look at line 4. echo 'It's time to stop writting errors "; Let’s examine the strings on this line. The intended string was: It’s time to stop writting errors. In PHP strings are quoted. Using the quotes, we see that our string i
phpMyAdminDB TOOL ブログ HOME » ブログ » PHPエラーの原因と対策 » エラーメッセージ » 「Parse error: syntax error, unexpected …」の行番号の箇所にエラーが見つからないとき 「Parse error: syntax error, unexpected …」の行番号の箇所にエラーが見つからないとき 投稿日 : 2012年7月7日 最終更新日時 : 2014年8月23日 作成者 : 西沢直木 カテゴリー : エラーメッセージ PHPスクリプトの入力を間違えると「Parse error」というエラーメッセージが表示されます。エラーメッセージの最後に、行番号が示されますが、必ずしもその行に原因があるとは限りらないので、その行だけを気にしすぎるとエラーの原因が把握できません。たとえば、次の例では、if文の閉じ忘れによって「Parse error:…… on line 14」が表示されますが、実際には7行目が原因です。 ○実行結果 「Parse error: syntax error, unexpected $end in C:\file……test.php on line 14」 定番のエラーメッセージを予習したい方におすすめの本
このようにエラーを修正するには慣れが必要です。実践で経験を積めば自然とエラーに対処するためのカンが身についてくるのですが、定番のエラーパターンを先に予習しておくのも1つの手です。 よくあるトラブルのパターンを予習しておきたい方にはこの本をおすすめします(私が書いた本ですが)。クイズ風にまとめてあります。実際に起きたエラーだと思ってクリアしていきましょう。 「PHP「直す力」養成ドリル」について詳しくはこちら スポンサーリンク こちらもどうぞ 詳細! PHP 7+MySQL 入門ノート 改訂新版 ゼロからわかるPHP超入門 WordPressテンプレートのifとwhile WordPressで予約システムを作ろう! PHPプログラミングの教科書 PHP「直す力」養成ドリル 土日でわかるPHPプログラミング教室 環境づくりからWebアプリが動くまでの2日間コース いまどきのアルゴリズムを使いこなす PHPプログラミング開発テクニック こちらも参考にしてください 関連記事 P