MySQL查詢不使用PHP的工作，但在查詢的phpmyadmin的SQL運行

Question

所以，這裏是我使用

<form action="upload.php" method="post" enctype="multipart/form-data"> 
<table border="2px"> 
<tr> 
<td>Date</td> 
<td>Day</td> 
<td>Height</td> 
<td>Weight</td> 
<td>Biceps(Open)</td> 
<td>Biceps(Curled)</td> 
<td>Photo</td> 
<td>Action</td> 
</tr> 
<tr> 
<td><input type="date" name = "date"></td> 
<td><select name="day"> 
<?php for ($x = 1; $x <= 100; $x++) { ?> 
    <option> 
    <?php echo $x; 
} ?></option> 
    </select> 
</td> 
<td><select name="height"> 
    <?php for ($x = 170; $x <= 180; $x++) { ?> 
    <option> 
    <?php echo $x; 
} ?></option> 
    </select> Centimeters &nbsp; 
</td> 
<td><select name="weight"> 
    <?php for ($x = 60; $x <= 80; $x++) { ?> 
    <option> 
    <?php echo $x; 
} ?></option> 
    </select> Kilograms &nbsp; 
</td> 
<td><select name="biceps_o"> 
    <?php for ($x = 10; $x <= 15; $x++) { ?> 
    <option> 
    <?php echo $x; 
} ?></option> 
    </select> Inches &nbsp; 
</td> 
<td><select name="biceps_c"> 
    <?php for ($x = 10; $x <= 15; $x++) { ?> 
    <option> 
    <?php echo $x; 
} ?></option> 
    </select> Inches &nbsp; 
</td> 
    Select image to upload: 
<td><input type="file" name="fileToUpload" id="fileToUpload"></td> 
<td><input type="submit" value="Add" name="submit"></td> 
</tr> 
</table> 
</form> 

這裏的HTML表單的upload.php的文件

<?php 
$target_dir = "images/"; 
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]); 
$uploadOk = 1; 
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION); 
// Check if image file is a actual image or fake image 
if(isset($_POST["submit"])) { 
    $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]); 
    if($check !== false) { 
     echo "File is an image - " . $check["mime"] . "."; 
     $uploadOk = 1; 
    } else { 
     echo "File is not an image."; 
     $uploadOk = 0; 
    } 
} 
// Check if file already exists 
if (file_exists($target_file)) { 
    echo "Sorry, file already exists."; 
    $uploadOk = 0; 
} 
// Check file size 
if ($_FILES["fileToUpload"]["size"] > 500000) { 
    echo "Sorry, your file is too large."; 
    $uploadOk = 0; 
} 
// Allow certain file formats 
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg" 
&& $imageFileType != "gif") { 
    echo "Sorry, only JPG, JPEG, PNG & GIF files are allowed."; 
    $uploadOk = 0; 
} 
// Check if $uploadOk is set to 0 by an error 
if ($uploadOk == 0) { 
    echo "Sorry, your file was not uploaded."; 
// if everything is ok, try to upload file 
} else { 
    if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) { 
     echo "The file ". basename($_FILES["fileToUpload"]["name"]). " has been uploaded."; 
    } else { 
     echo "Sorry, there was an error uploading your file."; 
    } 
} 
    $date = $_POST['date']; 
    $day = $_POST['day']; 
    $height = $_POST['height']; 
    $weight = $_POST['weight']; 
    $biceps_o = $_POST['biceps_o']; 
    $biceps_c = $_POST['biceps_c']; 

    echo $sql_query = "INSERT INTO gym (date, day, height, weight, biceps_o, biceps_c, image) VALUES('$date', '$day', '$height', '$weight', '$biceps_o', '$biceps_c', '$target_file')"; 
    if($sql_result = mysql_query($sql_query)) 
    echo "Data Successfully Added!"; 
    else 
    echo "Data Entry Failed!"; 
?> 

但是當我運行的代碼並上傳我的數據它提供了以下

文件是圖像 - 圖像/ jpeg.The文件 aid62387-728px交你的眼s-Step-2.jpg已上傳.INSERT INTO健身房（日期，日期，身高，體重，biceps_o，biceps_c，圖片） VALUES（'2016-08-15'，'1'，'170'，' 60'，'10'，'10'， 'images/aid62387-728px-Cross-Your-Eyes-Step-2.jpg'）數據輸入失敗！

好吧，圖像上傳完美，但查詢運行不正常。 但相同的查詢

INSERT INTO健身房（日期，星期，身高，體重，biceps_o，biceps_c，圖像） VALUES（ '2016年8月15日'， '1'， '170'，'60 」， '10'， '10'， '圖像/ aid62387-728px交你，眼睛步-2.JPG'）

工作完全正常，當我在phpMyAdmin的SQL運行它。可能是什麼問題？

Answer 1

我覺得還沒有包括烏爾數據庫連接，如PHP

Answer 2

您正在使用保留的mysql關鍵字（日期）。報價查詢如下：

$sql_query = "INSERT INTO gym (`date`, `day`, `height`, `weight`, `biceps_o`, `biceps_c`, `image`) VALUES('$date', '$day', '$height', '$weight', '$biceps_o', '$biceps_c', '$target_file')"; 

而且還添加一些逃脫角落找尋值（如mysql_real_escape_string）或開始使用PDO。