【发布时间】:2014-12-12 06:27:29
【问题描述】:
如何使用复选框检索以前保存在 mysql 中的数据以显示保存的内容?这是我的表格:
<?php
try {
$query = "select * from CONSULTA where user_id = '$user_id'";
$stmt = $conn->prepare( $query );
$stmt->execute();
while($row = $stmt->fetch()){
$sick = $row['sick'];
$user_id = $row['user_id'];
}
}catch(PDOException $exception){
echo "Error: " . $exception->getMessage();
}
?>
<form name="histo" id="histo" method="post">
<div class="row-fluid grid">
<label class="control-label"><b><?php echo $translate->__('Do you have any of the following symptoms'); ?>? :</b></label>
<div class="controls">
<label class="checkbox inline">
<div id="uniform-inlineCheckbox1" class="checker"><span><input style="opacity: 0;" id="inlineCheckbox1" value="1" name="sick[]" type="checkbox"<?php if ($sick == '1') {echo ' checked="checked"';} ?>></span></div> <?php echo $translate->__('Arrhythmias'); ?>
</label>
<label class="checkbox inline">
<div id="uniform-inlineCheckbox2" class="checker"><span><input style="opacity: 0;" id="inlineCheckbox2" value="2" name="sick[]" type="checkbox"<?php if ($sick == '2') {echo ' checked="checked"';} ?>></span></div> <?php echo $translate->__('Heart murmur'); ?>
</label>
<label class="checkbox inline">
<div id="uniform-inlineCheckbox3" class="checker"><span><input style="opacity: 0;" id="inlineCheckbox3" value="3" name="sick[]" type="checkbox"<?php if ($sick == '3') {echo ' checked="checked"';} ?>></span></div> <?php echo $translate->__('Stroke'); ?>
</label>
<label class="checkbox inline">
<div id="uniform-inlineCheckbox3" class="checker"><span><input style="opacity: 0;" id="inlineCheckbox4" value="4" name="sick[]" type="checkbox"<?php if ($sick == '4') {echo ' checked="checked"';} ?>></span></div> <?php echo $translate->__('Angina'); ?>
</label>
<label class="checkbox inline">
<div id="uniform-inlineCheckbox3" class="checker"><span><input style="opacity: 0;" id="inlineCheckbox5" value="5" name="sick[]" type="checkbox"<?php if ($sick == '5') {echo ' checked="checked"';} ?>></span></div> <?php echo $translate->__('Other'); ?>
</label>
</div>
</div>
<input type="hidden" name="user_id" value="<?php echo $_GET[user_id]; ?>" />
<?php if (!empty($visit)) { echo '<input type="hidden" name="action" value="edit" />'; } else { echo '<input type="hidden" name="action" value="crear" />';} ?>
<input type='submit' class='btn btn-primary' value='<?php $translate->__('Save History'); ?>' />
</form>
<div id="loading4" style="display:none;"><img src="img/ajax-loaders/loading4.gif" /></div>
<div id="oh"></div>
这是保存到数据库中的数据:
sick 1, 3, 5
但未显示复选框中的选中...错误在哪里?
【问题讨论】:
-
$sick的值是多少? -
您应该使用named parameters 准备语句,而不是仅仅填充任意用户数据。
$user_id之类的内容不应直接出现在您的查询中。 -
@u_mulder DB 中的值是这些= 1, 3, 5