Android WebView用postUrl加载页面后onBackPress无法加载上一页求解决方案

问题根源在于WebView默认的历史栈仅记录页面URL，不会保存POST请求的参数。当后退到POST加载的页面时，WebView无法自动重新发起带参数的POST请求，导致页面加载失败。下面提供两种可行的解决方案：

方案一：自定义历史栈维护请求信息 #

手动维护每个页面的请求类型（GET/POST）及POST参数，后退时根据记录重新发起请求。

1. 定义请求历史数据类 #

// 用Parcelable处理屏幕旋转等状态保存场景
@Parcelize
data class WebPageRecord(
    val url: String,
    val isPost: Boolean,
    val postData: ByteArray? = null
) : Parcelable {
    override fun equals(other: Any?): Boolean {
        if (this === other) return true
        if (javaClass != other?.javaClass) return false

        other as WebPageRecord

        if (url != other.url) return false
        if (isPost != other.isPost) return false
        if (postData != null) {
            if (other.postData == null) return false
            if (!postData.contentEquals(other.postData)) return false
        } else if (other.postData != null) return false

        return true
    }

    override fun hashCode(): Int {
        var result = url.hashCode()
        result = 31 * result + isPost.hashCode()
        result = 31 * result + (postData?.contentHashCode() ?: 0)
        return result
    }
}

2. 维护自定义历史栈并处理加载逻辑 #

在Activity/Fragment中：

private val pageHistory = mutableListOf<WebPageRecord>()
private lateinit var webView: WebView

// 封装POST加载方法
fun loadWithPost(url: String, postData: ByteArray) {
    pageHistory.add(WebPageRecord(url, true, postData))
    webView.postUrl(url, postData)
}

// 封装GET加载方法
fun loadWithGet(url: String) {
    pageHistory.add(WebPageRecord(url, false))
    webView.loadUrl(url)
}

3. 拦截WebView内部跳转，同步历史栈 #

webView.webViewClient = object : WebViewClient() {
    // 拦截页面内的GET跳转
    override fun shouldOverrideUrlLoading(view: WebView?, request: WebResourceRequest?): Boolean {
        request?.url?.toString()?.let {
            loadWithGet(it)
            return true
        }
        return super.shouldOverrideUrlLoading(view, request)
    }

    // 拦截页面内的POST表单提交
    override fun shouldInterceptRequest(
        view: WebView?,
        request: WebResourceRequest?
    ): WebResourceResponse? {
        request?.takeIf { it.method.equals("POST", ignoreCase = true) }?.let { req ->
            val url = req.url.toString()
            // Android 10+ 读取POST参数
            val postData = req.requestBody?.let { body ->
                val outputStream = ByteArrayOutputStream()
                body.writeTo(outputStream)
                outputStream.toByteArray()
            }
            postData?.let { data ->
                pageHistory.add(WebPageRecord(url, true, data))
            }
        }
        return super.shouldInterceptRequest(view, request)
    }
}

4. 重写onBackPressed处理后退逻辑 #

override fun onBackPressed() {
    if (webView.canGoBack() && pageHistory.size > 1) {
        // 移除当前页面记录
        pageHistory.removeLast()
        // 获取上一页的请求信息
        val lastPage = pageHistory.last()
        if (lastPage.isPost && lastPage.postData != null) {
            // 重新发起POST请求
            webView.postUrl(lastPage.url, lastPage.postData)
        } else {
            // GET请求直接后退
            webView.goBack()
        }
    } else {
        super.onBackPressed()
    }
}

5. 处理状态保存（可选） #

避免屏幕旋转时丢失历史栈：

override fun onSaveInstanceState(outState: Bundle) {
    super.onSaveInstanceState(outState)
    outState.putParcelableArrayList("web_page_history", ArrayList(pageHistory))
}

override fun onRestoreInstanceState(savedInstanceState: Bundle) {
    super.onRestoreInstanceState(savedInstanceState)
    savedInstanceState.getParcelableArrayList<WebPageRecord>("web_page_history")?.let {
        pageHistory.addAll(it)
    }
}

方案二：利用WebView的saveState与restoreState（局限性较大） #

WebView的saveState()方法会保存部分页面状态，但不保证能完整保存POST参数，仅适用于简单场景：

private var webViewState: Bundle? = null

override fun onSaveInstanceState(outState: Bundle) {
    super.onSaveInstanceState(outState)
    webViewState = webView.saveState(outState)
}

override fun onRestoreInstanceState(savedInstanceState: Bundle) {
    super.onRestoreInstanceState(savedInstanceState)
    webViewState?.let { webView.restoreState(it) }
}

此方法可靠性较低，推荐优先使用方案一。

内容的提问来源于stack exchange，提问作者anashabib365247