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PO
 

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    <p>You want to store your form data in your code.</p> <p>For this you have to make following changes in your code and database.</p> <h1>In Form</h1> <pre><code>note: Input field value should be relevant. With your existing html code your process.php will get data in this structure Array ( [textfield] =&gt; Array ( [childname] =&gt; dang [age] =&gt; 18 [parent_name] =&gt; doctor [contact_no] =&gt; 100 ) [checkbox] =&gt; Array ( [scratch] =&gt; checkbox [app_inventor] =&gt; checkbox [html] =&gt; checkbox [nov12] =&gt; checkbox [dec12] =&gt; checkbox ) [Submit] =&gt; Submit ) </code></pre> <p>So you need to modify your process.php</p> <p>Before that you have to modify structure of your table, follow these steps</p> <pre><code>1. Delete your existing table 2. Create new table DROP TABLE tblchildren ; CREATE TABLE tblchildren ( id INT AUTO_INCREMENT PRIMARY KEY, childname VARCHAR(30), age TINYINT, parent_name VARCHAR(30), contact_no VARCHAR(20), scratch ENUM('yes','no'), app_inventor ENUM('yes','no'), html ENUM('yes','no'), sesNov12Attnd ENUM('yes','no'), sesDec12Attnd ENUM('yes','no') ); </code></pre> <p>Since you are new to php so i have used basic function, I will suggest you to use switch from <strong>mysql to mysqli</strong></p> <h1>process.php</h1> <pre><code> &lt;?php $con=mysql_connect("hostname","username","pass"); $db=mysql_select_db('dbname', $con); //check form is submitted if(isset($_POST)){ //mysql_real_escape_string() prevents from sql injection. $childname= mysql_real_escape_string($_POST['textfield']['childname']); $age=mysql_real_escape_string($_POST['textfield']['age']); $parentName=mysql_real_escape_string($_POST['textfield']['parent_name']); $contactNo=mysql_real_escape_string($_POST['textfield']['contact_no']); $scratch=isset($_POST['checkbox']['scratch'])?'yes':'no'; $appInventor=isset($_POST['checkbox']['app_inventor'])?'yes':'no'; $html=isset($_POST['checkbox']['html'])?'yes':'no'; $nov12=isset($_POST['checkbox']['nov12'])?'yes':'no'; $dec12=isset($_POST['checkbox']['dec12'])?'yes':'no'; $sql="INSERT INTO tblchildren(childname, age, parent_name, contact_no, scratch,app_inventor,html, sesNov12Attnd, sesDec12Attnd ) VALUES ('$childname',$age,'$parentName','$contactNo','$scratch','$appInventor','$html','$nov12','$dec12')"; mysql_query($sql); }else{ echo "Form Not SUbmitted"; } ?&gt; </code></pre>
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    1. POHow to post array data from a form to mysql DB
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    1. POHow to post array data from a form to mysql DB
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    1. VO
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    1. COThanks so much - got so far and on submisison of form got this error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ',,no,no,yes,no,no)' at line 2. I'm running XAMPP with MySQL version 5.5.27...
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    2. COYou need to modify query as :- $sql="INSERT INTO tblchildren(childname, age, parent_name, contact_no, scratch,app_inventor,html, sesNov12Attnd, sesDec12Attnd ) VALUES ('$childname',$age,'$parentName','$contactNo','$scratch','$appInventor','$html','$nov12','$dec12')";
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      1. USDr_Dang
    3. COwhy would you use enum? use a bit(1) or a tinyint(1)
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      1. USmc_fish
 

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