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我已經創建了一個表單,它接受數據並將其插入到數據庫中。 onlick創建按鈕我希望函數檢查數據庫employee_id中的條目是否已經存在。如果存在,我希望它顯示的數據已經存在,你還是想插入,我在這可以任何人幫助我這個。形式是如何創建一個jquery函數來防止重複條目到數據庫
<form id="myForm" name="myForm" action='insert.php' method='post' >
<input type='hidden' name='st' value=0>
<table style="text-align:center; width:100%">
<tr>
<td style="text-align:right"><label>Select SE/AE:</label></td>
<td style="text-align:left">
<?php include("configs.php");
$sql = "SELECT DISTINCT seae FROM se_ae ";?>
<select name="seae">
<option value="" selected></option>
<?php foreach ($dbo->query($sql) as $row) { ?>
<option value="<?php echo $row['seae']; ?>">
<?php echo $row['seae']; ?></option>
<?php }?>
</select>
</td>
</tr>
<tr>
<td style="text-align:right"><label>Select Brand:</label></td>
<td style="text-align:left">
<?php //include("configs.php");
$sql = "SELECT DISTINCT `brand` FROM se_ae ";?>
<select name="brand">
<option value="" selected> </option>
<?php foreach ($dbo->query($sql) as $row) { ?>
<option value="<?php echo $row['brand']; ?>">
<?php echo $row['brand']; ?></option>
<?php }?>
</select>
</td>
</tr>
<tr>
<td style="text-align:right"><label>Select Territory:</label></td>
<td style="text-align:left">
<?php //include("configs.php");
$sql = "SELECT DISTINCT `territory` FROM se_ae ";?>
<select name="territory">
<option value="" selected></option>
<?php foreach ($dbo->query($sql) as $row) { ?>
<option value="<?php echo $row['territory']; ?>">
<?php echo $row['territory']; ?></option>
<?php }?>
</select>
</td>
</tr>
<tr>
<td style="text-align:right"><label for="name">Employee Name:</label></td>
<td style="text-align:left">
<input type="text" id="name" name="name"/>
</td>
</tr>
<tr>
<td style="text-align:right"><label for="number">Employee ID:</label></td>
<td style="text-align:left">
<input type="text" id="number" name="number" />
</td>
</tr>
<tr>
<td style="text-align:right"><label for="email"> Email:</label></td>
<td style="text-align:left">
<input type="text" id="email" name="email" />
</td>
</tr>
<tr>
<td style="text-align:right"><label for="contact"> Contact:</label></td>
<td style="text-align:left">
<input type="text" id="contact" name="contact" />
</td>
</tr>
<tr>
<td style="text-align:right"><label for="exist"> Exist:</label></td>
<td style="text-align:left">
<input type="radio" id="exist" name="exist" value="Active"/>Active
<input type="radio" id="exist" name="exist" value="NA"/>NA
</td>
</tr>
<tr>
<td style="text-align:right" class='swMntTopMenu'>
<input style="background-color:rgb(255,213,32)" name="Reset" type="reset" value="Reset">
</td>
<td style="text-align:left" class='swMntTopMenu'>
<input style="background-color:rgb(255,213,32)" name="submit" type="submit" value="Create">
</td>
</tr>
</table>
</form>
它可以達到使用jQuery的ajax.post .. –
@ user3113490如果你不介意你能解釋我我真的noob在這 –
jquery ajax後..就像一個mediator ....首先是你得到的輸入值,並將其傳遞到一個PHP文件,將做驗證...阿賈克斯是誰將通過的價值...像我說的像一箇中介。 –