Chrome和Opera在编辑记录时删除记录数据

问题描述 投票:0回答:1

我是php和mysql的新手,我正在使用像CMS这样的脚本,我的问题是我可以写新文章,但是当在chrome或opera上编辑它时,所有文章内容都没有显示用于编辑和删除,它不是使用firefox的情况,这里是编辑php脚本,

<?php echo link_tag('assets/themes/' . $this->selected_theme . '/'.UIKIT_VERSION.'/css/' . $this->session->userdata('selected-theme')); ?>
<script src="<?php echo base_url() ?>assets/themes/<?php echo $this->selected_theme . '/ckeditor/ckeditor.js'; ?>"></script>
<h4><i class="uk-icon-edit"></i> edit step </h4>
<hr class="uk-article-divider">
<?php $this->load->view('user/template/default/flash_data') ?>
<?php
$attributes = array(
    'method' => 'post',
    'class' => 'uk-form'
);
echo form_open_multipart('articles/edit/' . $step->step_id, $attributes) ?>
<div class="uk-form-controls">
    <label class="uk-form-label">
        <?php if ($step->step_photo_url) { ?>
            <img src="<?php echo $step->step_photo_url ?>" class="uk-thumbnail-small"/><br/>
        <?php } else { ?>
            <i class="uk-icon-image uk-icon-large"></i>
        <?php } ?>

    </label><br/>
    <input type="file" name="step_photo_url"/>
    <br/>
    <label class="uk-form-label uk-text-muted uk-text-bold">Or Enter a URL</label>
    <input class="uk-width-90 uk-form-large" type="text" placeholder="Image Url" name="step_photo_url"
           value="<?php echo set_value('step_photo_url', $step->step_photo_url, true) ?>"/>
    <?php echo form_error('step_photo_url') ?>
</div>
<div class="uk-form-controls uk-margin-top">
    <div class="uk-form-controls uk-margin-bottom">
        <input value="<?php echo $step->step_title ?>" name="step_title" class="uk-width-90 uk-form-large"
               type="text" placeholder="step title">
        <?php echo form_error('step_title'); ?>
    </div>
    <div class="uk-margin-bottom uk-form-controls">
            <textarea name="step_description" rows="4" class="uk-width-90 uk-form-large ckeditor"
                      placeholder="Enter Description"><?php echo set_value('step_description', $step->step_description,
                    true) ?></textarea>
        <?php echo form_error('step_description'); ?>
    </div>
    <a href="<?php echo site_url('articles/delete/' . $step->step_id) ?>"
       class="uk-button uk-button-danger uk-float-left">Delete</a>
    <button class="uk-button uk-button-success uk-float-right">Save</button>
</div>
</form>
php mysql google-chrome
1个回答
0
投票

这是一个难以猜测的问题,虽然这听起来像我第一次开始使用PHP和数据库时犯的错误。那就是使用htmlspecialchars($value,ENT_QUOTES)转换值字段中的输出。如果你有这个值:

$content = '"What are you talking about?" she asked, "that is ridiculous!"';

然后你回到输入字段进行编辑:

<input name="test" value="<?php echo $content ?>" />

html输出是:

<input name="test" value=""What are you talking about?" she asked, "that is ridiculous!"" />

它有两组引号,然后大多数浏览器会将该字段显示为空白。然后,当您单击更新时,它会清除您之前拥有的内容。要解决这个问题,你应该使用:

<input name="test" value="<?php echo htmspecialchars($content,ENT_QUOTES) ?>" />

然后在现场给你这个输出:

"What are you talking about?" she asked, "that is ridiculous!"

但是html看起来像这样:

<input name="test" value="&quot;What are you talking about?&quot; she asked, &quot;that is rediculous!&quot;" />

根据您的描述并根据个人经验,这只是在黑暗中刺伤。您将必须显示您的HTML或一些PHP值以获得更清晰的图片。

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