I got trouble uploading image and saving records in Mysql database using PHP, can somebody help me?
Code:
<?php
ini_set('mysql.connect_timeout', 300);
ini_set('default_socket_timeout', 300);
?>
<html>
<body>
<form method="post" enctype="multipart/form-data">
</br>
<input type="text" name="dbname"/>
<input type="file" name="dbimage"/>
<br> <br>
<input type="submit" name"submit" value"Upload"/>
</form>
<?php
if(isset($_POST['submit'])){
if(getimage($_FILES['dbimage']['tmp_name']) == FALSE) {
echo "Please select an image";
} else {
$dbimage = addcslashes($_FILES['dbimage']['tmp_name']);
$dbname = addcslashes($_FILES['dbimage']['dbname']);
$dbimage = file_get_contents($image);
$dbimage = base64_encode($dbimage);
saveimage($dbname, $dbimage);
}
function saveimage() {
$con = mysql_connect("localhost", "root", "");
mysql_select_db("db_test", $con);
$qry = "insert into table1 (dname,dpic) values ('$dbname','$dbimage')";
$result = mysql_query($qry, $con);
if ($result){
echo "Image uploaded.";
} else {
echo " Image not uploaded.";
}
}
}?>
</body>
</html>
你必须给函数 saveimage() 变量 $dbname 和 $dbimage ... 尽管如此,请使用 PDO ...
You have missed parameters in function saveimage(). It should be like this: function saveimage($dbname,$dbimage)
I think that you forgot to pass the params to your function
function saveimage($dbname,$dbimage){...}
Try to declare your function outside of your condition.
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