我在保存到數據庫時遇到了這個選擇選項問題這是我的代碼<div class="col-sm-2 control-label no-padding-right"> <div id="loantype"> <select class="form-group loan_type" id="loan_type form-field-2" name="loan_type" value="<?php echo set_value('loan_type');?>" required> <option value="0">Select Loan Type</option> <option value="Personal">Personal</option> <option value="Business">Business</option> </select> </div> </div> <div class="col-sm-2 control-label no-padding-right"> <div class="personalloan" id="personalloan" style="display:none"> <select class="form-group loan_amount" id="loan_amount form-field-2" name="loan_amount" value="<?php echo set_value('loan_amount');?>" required> <option value="0">Select Loan Amount</option> <option value="10000">P10,000</option> <option value="20000">P20,000</option> <option value="30000">P30,000</option> </select> </div>Business如果我選擇將出現的話也是如此businessloan。但問題是只有選擇選項businessloan中的數據被保存在數據庫中，但是當我在personalloan其始終為 0 上選擇某些內容時。我這樣保存'loan_type' => $this->input->post('loan_type'),'loan_amount' => $this->input->post('loan_amount'),有什么問題？
2 回答
子衿沉夜
TA貢獻1828條經驗 獲得超3個贊
讓我們從一些問題開始。
您
id在多個元素上有相同的內容。id's'在文檔中必須是唯一的。您
name在兩個選擇上都有相同的選擇，這將使一個在表單發布時覆蓋另一個
因此，讓我們從更改名稱和 ID 開始：
<!-- The personal select -->
<select class="form-group loan_amount" id="loan_amount_personal" name="loan_amount_personal" required>
<!-- The business select -->
<select class="form-group loan_amount" id="loan_amount_business" name="loan_amount_business" required>
如果有的話，您可能需要用新的 id 更新 javascript 中的任何引用。
然后，當表單發布時，您會根據類型獲得所需的選擇：
$amount = null;
if ($this->input->post('loan_type') === 'Personal') {
$amount = $this->input->post('loan_amount_personal');
} else {
$amount = $this->input->post('loan_amount_business');
}
// Or for short:
$amount = $this->input->post('loan_type') === 'Personal'
? $this->input->post('loan_amount_personal')
: $this->input->post('loan_amount_business');
然后，在構建數組時：
'loan_type' => $this->input->post('loan_type'),
'loan_amount' => $amount,
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