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POhow do i store the value of stars rating in mysql database
 

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  1. POhow do i store the value of stars rating in mysql database
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    <p>Hi I have form that contains Ratings, Name, email, and comments. I am able to insert the user input data for name, email and comments. But don't know how to store the star ratings in database. Anyone please help me. thanks</p> <pre><code> &lt;?php if(isset($_POST['submit'])) { $name = $_POST['name']; $email = $_POST['email']; $comments = $_POST['comments']; $ratings = $_POST['ratings']; $link = mysqli_connect("localhost", "root", "", "imakr"); /* check connection */ if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit(); } $res = mysqli_query($link, "insert into imakr.customer_review(name, email, comments, ratings) values('$name','$email','$comments', '$ratings')"); if($res) { echo "Your feedback is saved"; } else { echo " OOPs!! there is some error. Please check the fields"; } } ?&gt; &lt;form id="customer_review" name="cust_rev"action="" method="post" onsubmit="return validate()"&gt; &lt;table width="535" border="0"&gt; &lt;tr&gt; &lt;td&gt;Rate This Product: &lt;/td&gt; &lt;td&gt; &lt;span id="rateStatus"&gt;Rate Me...&lt;/span&gt; &lt;span id="ratingSaved"&gt;Rating Saved!&lt;/span&gt; &lt;div id="rateMe" title="Rate Me..."&gt; &lt;a onclick="rateIt(this)" id="_1" title="Poor" onmouseover="rating(this)" onmouseout="off(this)"&gt;&lt;span class="ratings"&gt;1&lt;/span&gt;&lt;/a&gt; &lt;a onclick="rateIt(this)" id="_2" title="Not Bad" onmouseover="rating(this)" onmouseout="off(this)"&gt;&lt;span class="ratings"&gt;2&lt;/span&gt;&lt;/a&gt; &lt;a onclick="rateIt(this)" id="_3" title="Pretty Good" onmouseover="rating(this)" onmouseout="off(this)"&gt;&lt;span class="ratings"&gt;3&lt;/span&gt;&lt;/a&gt; &lt;a onclick="rateIt(this)" id="_4" title="Excellent" onmouseover="rating(this)" onmouseout="off(this)"&gt;&lt;span class="ratings"&gt;4&lt;/span&gt;&lt;/a&gt; &lt;a onclick="rateIt(this)" id="_5" title="Marvellous" onmouseover="rating(this)" onmouseout="off(this)"&gt;&lt;span class="ratings"&gt;5&lt;/span&gt;&lt;/a&gt; &lt;/div&gt; &lt;/td&gt; &lt;/tr&gt; &lt;tr&gt; &lt;td width="129"&gt;&lt;span class="titles"&gt;Name&lt;/span&gt;&lt;span class="star"&gt;*&lt;/span&gt;:&lt;/td&gt; &lt;td width="396"&gt;&lt;label for="name"&gt;&lt;/label&gt; &lt;input type="text" name="name" id="name" /&gt;&lt;/td&gt; &lt;/tr&gt; &lt;tr&gt; &lt;td&gt;&lt;span class="titles"&gt;Email&lt;/span&gt;&lt;span class="star"&gt;*&lt;/span&gt;:&lt;/td&gt; &lt;td&gt;&lt;label for="email"&gt;&lt;/label&gt; &lt;input type="text" name="email" id="email" /&gt;&lt;/td&gt; &lt;/tr&gt; &lt;tr&gt; &lt;td height="61"&gt;Comments:&lt;/td&gt; &lt;td&gt;&lt;label for="comments"&gt;&lt;/label&gt; &lt;textarea name="comments" id="comments" cols="45" rows="5" onchange="maxlength('comments', 500)"&gt;&lt;/textarea&gt;&lt;/td&gt; &lt;/tr&gt; &lt;tr&gt; &lt;td&gt;&amp;nbsp;&lt;/td&gt; &lt;td&gt;&lt;input type="submit" name="submit" id="submit" value="Submit" /&gt;&lt;/td&gt; &lt;/tr&gt; &lt;/table&gt; &lt;p&gt;&amp;nbsp;&lt;/p&gt; &lt;/form&gt; </code></pre>
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    1. USUjjwal Chowdary
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    1. COImmediately stop using this code. It is vulnerable to SQL injection. You're using an API that is deprecated. Learn about [*prepared statements*](http://j.mp/T9hLWi) instead, and use [PDO](http://php.net/pdo) or [MySQLi](http://php.net/mysqli) - [this article](http://j.mp/QEx8IB) will help you decide which.
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      1. USKermit
    2. CO@Polish Prince Sorry, can you tell me why? I am a beginner.
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      1. USUjjwal Chowdary
    3. COyes it is, I would recommend using mysqli, check out this link http://php.net/manual/de/book.mysqli.php
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      1. USmakim
 

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