本文介绍了PDO-将Blob图片插入MySQL的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一些问题
这是我的表格:
<form action="contact_ajouter_verif.php" method="post" name="ajoutContact" enctype="multipart/form-data" >
<fieldset>
<label>Nom :</label> <input size="30%" type="text" placeholder="" name="nom" />
<label>Numéro :</label> <input size="30%" type="number" placeholder="" name="num" />
<label>Image au format png :</label><input type="file" name="img" />
</fieldset>
<input name="submit" type="submit" value="Ajouter"/>
</form>
这是我的pdo接收器页面:
And this is my pdo receiver page:
<?php
include('../inc/connexion.inc.php');
include('session.php');
$nom = $_POST['nom'];
$num = $_POST['num'];
$img = $_FILES['img'];
$pseudo = $user_check. "_contact";
$rqt1= "INSERT INTO $pseudo(CTC_NOM, CTC_NUMERO, CTC_IMG) VALUES(:nom, :num, :img)";
$result1 =$cnxpdo->prepare($rqt1);
$result1->execute(array(
'nom' => "$nom",
'num' => "$num",
'img' => "$img" //line 17
));
?>
我真的不明白我在做什么错,请提供解决方案的人:)
I really don't understand what I'm doing wrong, please if someone have the solution :)
推荐答案
最后找到了和我一样的人(不太可能,但我们永远不知道...):
Finally found for those in the same case than me (unlikely but we never know...) :
<?php
include('../inc/connexion.inc.php');
include('session.php');
$nom = $_POST['nom'];
$num = $_POST['num'];
$img =addslashes(file_get_contents ($_FILES['img']['tmp_name']));
$pseudo = $user_check. "_contact";
$rqt1= "INSERT INTO $pseudo(CTC_NOM, CTC_NUMERO, CTC_IMG) VALUES(:nom, :num, :img)";
$result1 =$cnxpdo->prepare($rqt1);
$result1->execute(array(
'nom' => "$nom",
'num' => "$num",
'img' => "$img"
));
?>
感谢您的帮助.
这篇关于PDO-将Blob图片插入MySQL的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!