Question

我目前正在开发一个PHP脚本来将图像上传到mysql数据库，我在代码的第13行收到语法错误，指出了意外的T_IF。我对PHP比较陌生，我正在努力找出导致它的原因。代码如下;

    <head>
        <title>File Upload To Database</title>
    </head>
    <body>
        <form action="index.php" method="POST" enctype="multipart/form-data">
            <input type="file" name="image"><Input type="submit" name="submit" value="Upload">
        </form>

        <?php>
        if(isset($_POST['submit']))
            {
                mysql_connect("localhost","appuser1","******");
                mysql_select_db("PHP_Sig_test_app");

                $imageName = mysql_real_escape_string($FILES["image"]["name"]);
                $imageData = mysql_real_escape_string(file_get_contents($_FILES["image"]["tmp_name"]));
                $imageType = mysql_real_escape_string($FILES["image"]["type""]);

                if(substr($imageType,0,5) == "image)
                {
                    mysql_query("INSERT INTO 'Data' Values('','$imageName','$imageData')";
                    echo "image uploaded";
                }
                else
                {
                    echo "Only Images can be Inserted"
                }    
            }
        ?>
    </body>
</html>

由于

Answer 1

<?php>应为<?php

导致语法错误

Answer 2

删除此行中的右括号：

<?php>

这会启动一个带有>符号的新命令（这使得没有意义），因此> if将失败，这就是您的错误原因。

你的第二个if语句也缺少引用。

if(substr($imageType,0,5) == "image)

应该是

if(substr($imageType,0,5) == "image")

Answer 3

有一些语法错误 -

<?php //removed >

    if(isset($_POST['submit']))
        {
            mysql_connect("localhost","appuser1","weeble180276");
            mysql_select_db("PHP_Sig_test_app");

            $imageName = mysql_real_escape_string($FILES["image"]["name"]);
            $imageData = mysql_real_escape_string(file_get_contents($_FILES["image"]["tmp_name"]));
            $imageType = mysql_real_escape_string($FILES["image"]["type"]); //removed "

            if(substr($imageType,0,5) == "image") //added "
            {
                mysql_query("INSERT INTO 'Data' Values('','$imageName','$imageData')";
                echo "image uploaded";
            }
            else
            {
                echo "Only Images can be Inserted"; // added ;
            }
        }
    ?>

意外的T_IF语法错误

3 个答案: