I have a website and it uses a lytebox to show some images and other websites on top of a current website. This all works fine, but I have transferred the data into a MYSQL database and now the class of the line does not work.
我有一个网站,它使用lytebox在当前网站上显示一些图像和其他网站。这一切都运行正常,但我已将数据传输到MYSQL数据库,现在该行的类不起作用。
This is what I had previously and this works:
这就是我以前的工作原理:
<div id="photo01">
<a href="images/strayKatts/photo01.JPG" title="" class="thickbox">
<img src="images/strayKatts/photo01.JPG" alt="Stray Katts Creations"
width="150px" height="113px" border="0"/>
</a>
</div>
but because I am getting more and more data in the directory I have transferred my data into a mysql database to maintain a lot easier.
但是因为我在目录中获得越来越多的数据,所以我将数据传输到mysql数据库中以保持更轻松。
I can pull the data and everything, it is just not being displayed probably and I currently have this:
我可以提取数据和所有内容,它可能没有显示,我目前有这个:
<div id="photo01">
<?php
$category_set = mysql_query("SELECT * FROM currentmarkets
WHERE id = '3' AND visible = '1';", $dbconnect);
if (!$category_set) {
die("Database query failed: " . mysql_error());
}
while ($titleItem = mysql_fetch_array($category_set)) {
echo "<a href='{$titleItem["imageLink01"]}' title='' class='thickbox'><img src='{$titleItem["imageSrc01"]}' alt='{$titleItem["imageAlt01"]}' width='150px' height='113px' border='0'/>";}
?></div>
I am sure it has something to do with the way I am using my '' and my "" not sitting right.
我确信这与我使用''和我'“的方式有关。
Any more advice / help?
还有什么建议/帮助吗?
EDIT!!
Sorry maybe I am not doing it right, but I have tried all the answer and none of them change anything even though they seem right - this is confusing:-)
对不起,也许我做得不对,但我已经尝试了所有的答案,即使他们看起来是对的,也没有任何改变 - 这很令人困惑:-)
oh and sorry I forgot that - I'll never do that again:-)
哦,对不起,我忘记了 - 我永远不会再这样做了:-)
5 个解决方案
#1
2
Try like this, using concatenation. Also try to just print the array fetched to see if it's a problem of your embedded html in php or to check if the query is not ok.
尝试这样,使用连接。还试着打印所提取的数组,看看它是否是你在php中嵌入html的问题,或者检查查询是否正常。
<div id="photo01">
<?php
$category_set = mysql_query("SELECT * FROM currentmarkets WHERE id = '3' AND visible = '1';", $dbconnect);
if (!$category_set) {
die("Database query failed: " . mysql_error());
}
while ($titleItem = mysql_fetch_array($category_set)) {
echo "<a href='".$titleItem["imageLink01"]."' title='' class='thickbox'><img src='".$titleItem["imageSrc01"]."' alt='".$titleItem["imageAlt01"]."' width='150px' height='113px' border='0'/>";}
?></div>
#2
2
change:
echo "<a href='{$titleItem["imageLink01"]}' title='' class='thickbox'><img src='{$titleItem["imageSrc01"]}' alt='{$titleItem["imageAlt01"]}' width='150px' height='113px' border='0'/>";}
to:
echo "<a href='" . $titleItem["imageLink01"] . "' title='' class='thickbox'><img src='" . $titleItem["imageSrc01"] . "' alt='" . $titleItem["imageAlt01"] . "' width='150px' height='113px' border='0'/></a>";}
#3
1
In your while
loop:
在你的while循环中:
while ($titleItem = mysql_fetch_array($category_set)) {
echo "<a href='".$titleItem["imageLink01"]."' title='' class='thickbox'>";
echo "<img src='".$titleItem["imageSrc01"]."' alt='".$titleItem["imageAlt01"]."'";
echo " width='150px' height='113px' border='0'/></a>";
}
You forgot the </a>
closing tag, and also need to just use concatenation to echo
your vars from the query results.
您忘记了结束标记,还需要使用连接来从查询结果中回显您的变量。
#4
0
This should probably work, I don't see any other mistake into your code:
这应该可行,我没有看到你的代码中的任何其他错误:
echo "<a href='" . $titleItem["imageLink01"] . "' title='' class='thickbox'><img src='" . $titleItem["imageSrc01"] . "' alt='" . $titleItem["imageAlt01"]. "' width='150px' height='113px' border='0'/></a>";
#5
0
Uses a heredoc when you've got a long mix of HTML and PHP variables:
当你有很多混合的HTML和PHP变量时,使用heredoc:
while ($titleItem = mysql_fetch_array($category_set)) {
echo <<<EOL
<a href="{$titleItem['imageLink01']}" title="" class="thickbox">
<img src="{$titleItem['imageSrc01']}" alt="{$titleItem['imageAlt01']}" width="150px" height="113px" border="0" />
</a>
EOL;
}
#1
2
Try like this, using concatenation. Also try to just print the array fetched to see if it's a problem of your embedded html in php or to check if the query is not ok.
尝试这样,使用连接。还试着打印所提取的数组,看看它是否是你在php中嵌入html的问题,或者检查查询是否正常。
<div id="photo01">
<?php
$category_set = mysql_query("SELECT * FROM currentmarkets WHERE id = '3' AND visible = '1';", $dbconnect);
if (!$category_set) {
die("Database query failed: " . mysql_error());
}
while ($titleItem = mysql_fetch_array($category_set)) {
echo "<a href='".$titleItem["imageLink01"]."' title='' class='thickbox'><img src='".$titleItem["imageSrc01"]."' alt='".$titleItem["imageAlt01"]."' width='150px' height='113px' border='0'/>";}
?></div>
#2
2
change:
echo "<a href='{$titleItem["imageLink01"]}' title='' class='thickbox'><img src='{$titleItem["imageSrc01"]}' alt='{$titleItem["imageAlt01"]}' width='150px' height='113px' border='0'/>";}
to:
echo "<a href='" . $titleItem["imageLink01"] . "' title='' class='thickbox'><img src='" . $titleItem["imageSrc01"] . "' alt='" . $titleItem["imageAlt01"] . "' width='150px' height='113px' border='0'/></a>";}
#3
1
In your while
loop:
在你的while循环中:
while ($titleItem = mysql_fetch_array($category_set)) {
echo "<a href='".$titleItem["imageLink01"]."' title='' class='thickbox'>";
echo "<img src='".$titleItem["imageSrc01"]."' alt='".$titleItem["imageAlt01"]."'";
echo " width='150px' height='113px' border='0'/></a>";
}
You forgot the </a>
closing tag, and also need to just use concatenation to echo
your vars from the query results.
您忘记了结束标记,还需要使用连接来从查询结果中回显您的变量。
#4
0
This should probably work, I don't see any other mistake into your code:
这应该可行,我没有看到你的代码中的任何其他错误:
echo "<a href='" . $titleItem["imageLink01"] . "' title='' class='thickbox'><img src='" . $titleItem["imageSrc01"] . "' alt='" . $titleItem["imageAlt01"]. "' width='150px' height='113px' border='0'/></a>";
#5
0
Uses a heredoc when you've got a long mix of HTML and PHP variables:
当你有很多混合的HTML和PHP变量时,使用heredoc:
while ($titleItem = mysql_fetch_array($category_set)) {
echo <<<EOL
<a href="{$titleItem['imageLink01']}" title="" class="thickbox">
<img src="{$titleItem['imageSrc01']}" alt="{$titleItem['imageAlt01']}" width="150px" height="113px" border="0" />
</a>
EOL;
}