0

如何将图片路径存入数据库并在上传后显示？

<?php
$sub=0;
ini_set( "display_errors", 0);
if(isset($_REQUEST['submited'])) {
// your save code goes here

$allowedExts = array("jpg", "jpeg", "gif", "png");
$extension = end(explode(".", $_FILES["file"]["name"]));
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/png")
|| ($_FILES["file"]["type"] == "image/pjpeg"))
&& ($_FILES["file"]["size"] < 2097152)
&& in_array($extension, $allowedExts))
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
}
else
{
echo "";
if (file_exists("images/" . $_FILES["file"]["name"]))
{
echo "<font size='4' color='red'><b>We are sorry, the file you trying to upload already exists.</b></font>";
  }

else
{
move_uploaded_file($_FILES["file"]["tmp_name"],
"images/" . $_FILES["file"]["name"]);
$sub= 1;
echo "<font size='7' color='white'><b> Success! Your photo has been uploaded.</b></font>";

}

}
}
else
{
echo "<font size='4' color='red'><b>We are sorry, the file you trying to upload is not an image or it exceeds 2MB in size.</b></font><br><font color='blue'><i>Only images under size of 2MB are allowed</i></font>.";
}
}

?>
<form action="" method="post" enctype="multipart/form-data">
<input type="hidden" name="submited" value="true" />

<?php
ini_set( "display_errors", 0);
if($sub==0)
{
?> 
<label size="16" for="file">Choose Photo:</label>
<input id="shiny" type="file" name="file" onchange="file_selected = true;">
<input id="shiny" type="submit" value="Upload" name="submit">
<?php
}
?>

</form>

这是数据库信息...在将图像路径插入数据库后如何显示图片？我试过VALUES ('$_FILES["file"]["name"]')";了，但这似乎不起作用..

<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("simple_login", $con);

$sql="INSERT INTO photo (photo)
VALUES
('$_FILES["file"]["name"]')";

if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}

mysql_close($con);
?>
4

2 回答 2

1 
"INSERT INTO photo (photo) VALUES ('{$_FILES["file"]["name"]}')"

那应该行得通。要在字符串中使用关联数组，您必须将其包裹在大{ }括号 ( ) 中。

我想说的 3 点与具体问题无关：

1：在放入数据库之前，您应该始终清理用户输入。所以你应该做的是：

"INSERT INTO photo (photo) VALUES ('" . mysql_real_escape_string($_FILES["file"]["name"]) . "')"

或使用带有 mysqli 或 pdo 的准备好的语句。

2：如果你只是在数据库中存储一个文件列表，那有什么意义呢？为什么不只遍历存储它们的目录呢？

3：mysql_*功能贬值，应考虑使用mysqli或pdo

于 2013-01-03T22:09:31.803 回答
0

我刚刚使用 Mysqli 解决了它，所以我也可以防止 sql 注入.....谢谢你们的帮助......

<?php
$sub=0;
ini_set( "display_errors", 0);
if(isset($_REQUEST['submited'])) {

// your save code goes here

$allowedExts = array("jpg", "jpeg", "gif", "png");
$extension = end(explode(".", $_FILES["file"]["name"]));
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/png")
|| ($_FILES["file"]["type"] == "image/pjpeg"))
&& ($_FILES["file"]["size"] < 2097152)
&& in_array($extension, $allowedExts))
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
}
else
{
echo "";
if (file_exists("images/" . $_FILES["file"]["name"]))
{
echo "<font size='4' color='red'><b>We are sorry, the file you trying to upload already exists.</b></font>";
  }

else
{
move_uploaded_file($_FILES["file"]["tmp_name"],
"images/" . $_FILES["file"]["name"]);
$sub= 1;
$mysqli = new mysqli("localhost", "root", "", "simple_login");

// TODO - Check that connection was successful.

$photo= $_FILES["file"]["name"];

$stmt = $mysqli->prepare("INSERT INTO photo (photo) VALUES (?)");

// TODO check that $stmt creation succeeded

// "s" means the database expects a string
$stmt->bind_param("s", $photo);

$stmt->execute();

$stmt->close();

$mysqli->close();

echo "<font size='7' color='white'><b> Success! Your photo has been uploaded.</b></font>";
}

}
}
else
{
echo "<font size='4' color='red'><b>We are sorry, the file you trying to upload is not an image or it exceeds 2MB in size.</b></font><br><font color='blue'><i>Only images under size of 2MB are allowed</i></font>.";
}
}

?>
<form action="" method="post" enctype="multipart/form-data">
<input type="hidden" name="submited" value="true" />


<?php
ini_set( "display_errors", 0);
if($sub==0)
{
?> 
<label size="16" for="file">Choose Photo:</label>
<input id="shiny" type="file" name="file" onchange="file_selected = true;">
<input id="shiny" type="submit" value="Upload" name="submit">
<?php
}
?>


</form>
</div>
于 2013-01-04T00:02:41.560 回答